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Question 1
Sulfur dioxide reacts with oxygen to form sulfur trioxide according to the balanced equation: $2 SO_{2} (g) + O_{2} (g) \to 2 SO_{3} (g)$ A chemist combines $60.0 cm^{3}$ of $SO_{2}$ with $50.0 cm^{3}$ of $O_{2}$ in a sealed vessel. Assuming the reaction goes to completion, what is the total volume of gas remaining? (All volumes are measured at the same temperature and pressure.)
No clue? Show me the answer
Correct answerCorrect!Incorrect
C $80.0 cm^{3}$
Step-by-step walkthrough
Choose a solution method
Method #1Approach 1
Step 1: Identify the mole ratio and limiting reactant
The balanced equation is $2 SO_{2} + O_{2} \to 2 SO_{3}$ . The mole ratio is $SO_{2} : O_{2} = 2 : 1$ . At constant temperature and pressure, volume ratios equal mole ratios.
Step 2: Determine which reactant is limiting
To consume all $60.0 cm^{3}$ of $SO_{2}$ , we need $\frac{60.0}{2} = 30.0 cm^{3}$ of $O_{2}$ . We have $50.0 cm^{3}$ of $O_{2}$ , which is more than enough. So $SO_{2}$ is the limiting reactant.
Step 3: Calculate volumes consumed and produced
All $60.0 cm^{3}$ of $SO_{2}$ is consumed. $O_{2}$ consumed $= 30.0 cm^{3}$ . $O_{2}$ remaining $= 50.0 - 30.0 = 20.0 cm^{3}$ . $SO_{3}$ produced $= 60.0 cm^{3}$ (ratio $2 : 2 = 1 : 1$ with $SO_{2}$ ).
Step 4: Calculate total gas volume remaining
Total volume $= V (SO_{3}) + V (O_{2} excess) = 60.0 + 20.0 = 80.0 cm^{3}$ .
Method #2Approach 2
Step 1: Identify what the question asks
We need the total gas volume after complete reaction, accounting for what is consumed and what remains (including excess reactant and product).
Step 2: Eliminate $110.0 \text{ cm}^3$
This would imply no gas is consumed at all — clearly incorrect since the reaction converts $SO_{2}$ and $O_{2}$ into $SO_{3}$ , and the total moles of gas decrease.
Step 3: Eliminate $60.0 \text{ cm}^3$
This would suggest only the $SO_{3}$ product remains with no excess $O_{2}$ . But $SO_{2}$ is the limiting reactant, so $O_{2}$ is in excess — $20.0 cm^{3}$ of $O_{2}$ remains unreacted.
Step 4: Eliminate $50.0 \text{ cm}^3$
This is simply the initial volume of $O_{2}$ , which does not correspond to any meaningful calculation of remaining gas after the reaction.
Step 5: Select the correct answer
$SO_{3}$ produced $= 60.0 cm^{3}$ ; excess $O_{2} = 20.0 cm^{3}$ . Total $= 80.0 cm^{3}$ . The correct answer is $80.0 cm^{3}$ .
Question 2
The thermite reaction used in welding is: $2 Al (s) + Fe_{2} O_{3} (s) \to Al_{2} O_{3} (s) + 2 Fe (l)$ Using $A_{r}$ values: Al = 27, Fe = 56, O = 16, which expression correctly calculates the atom economy for producing iron ( $Fe$ )?
No clue? Show me the answer
Correct answerCorrect!Incorrect
A $\frac{112}{102 + 112} \times 100$
Step-by-step walkthrough
Choose a solution method
Method #1Approach 1
Step 1: Recall the atom economy formula
$Atom economy = \frac{M _{r} ( desired product )}{\sum M _{r} ( all products )} \times 100$
Step 2: Identify all products and their molar masses
The two products are $Al_{2} O_{3}$ and $2 Fe$ . $M_{r} (Al_{2} O_{3}) = 2 (27) + 3 (16) = 54 + 48 = 102$ . $M_{r} (2 Fe) = 2 \times 56 = 112$ .
Step 3: Identify the desired product
The desired product is iron (Fe). The coefficient is 2 in the balanced equation, so the relevant molar mass is $2 \times 56 = 112 g mol^{- 1}$ .
Step 4: Write the correct expression
$Atom economy = \frac{112}{102 + 112} \times 100 = \frac{112}{214} \times 100 \approx 52.3%$
Method #2Approach 2
Step 1: Identify what is needed
Atom economy uses the sum of molar masses of ALL products in the denominator, and the molar mass of the desired product only in the numerator — both calculated using stoichiometric coefficients from the balanced equation.
Step 2: Eliminate $\frac{56}{214} \times 100$
This uses the molar mass of a single Fe atom (56) rather than the 2 mol of Fe shown in the equation. The coefficient must be included, so numerator should be $2 \times 56 = 112$ .
Step 3: Eliminate $\frac{102}{214} \times 100$
This uses $Al_{2} O_{3}$ (the unwanted by-product, $M_{r} = 102$ ) as the desired product in the numerator. The desired product is Fe, not $Al_{2} O_{3}$ .
Step 4: Eliminate $\frac{112}{102} \times 100$
This omits $Al_{2} O_{3}$ from the denominator entirely. The denominator must include all products: $102 + 112 = 214$ , not just one product.
Step 5: Select the correct expression
$\frac{112}{102 + 112} \times 100$ correctly places the molar mass of 2 Fe in the numerator and the sum of molar masses of all products in the denominator.

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Question 1
Sulfur dioxide reacts with oxygen to form sulfur trioxide according to the balanced equation: $2 SO_{2} (g) + O_{2} (g) \to 2 SO_{3} (g)$ A chemist combines $60.0 cm^{3}$ of $SO_{2}$ with $50.0 cm^{3}$ of $O_{2}$ in a sealed vessel. Assuming the reaction goes to completion, what is the total volume of gas remaining? (All volumes are measured at the same temperature and pressure.)
No clue? Show me the answer
Correct answerCorrect!Incorrect
C $80.0 cm^{3}$
Step-by-step walkthrough
Choose a solution method
Method #1Approach 1
Step 1: Identify the mole ratio and limiting reactant
The balanced equation is $2 SO_{2} + O_{2} \to 2 SO_{3}$ . The mole ratio is $SO_{2} : O_{2} = 2 : 1$ . At constant temperature and pressure, volume ratios equal mole ratios.
Step 2: Determine which reactant is limiting
To consume all $60.0 cm^{3}$ of $SO_{2}$ , we need $\frac{60.0}{2} = 30.0 cm^{3}$ of $O_{2}$ . We have $50.0 cm^{3}$ of $O_{2}$ , which is more than enough. So $SO_{2}$ is the limiting reactant.
Step 3: Calculate volumes consumed and produced
All $60.0 cm^{3}$ of $SO_{2}$ is consumed. $O_{2}$ consumed $= 30.0 cm^{3}$ . $O_{2}$ remaining $= 50.0 - 30.0 = 20.0 cm^{3}$ . $SO_{3}$ produced $= 60.0 cm^{3}$ (ratio $2 : 2 = 1 : 1$ with $SO_{2}$ ).
Step 4: Calculate total gas volume remaining
Total volume $= V (SO_{3}) + V (O_{2} excess) = 60.0 + 20.0 = 80.0 cm^{3}$ .
Method #2Approach 2
Step 1: Identify what the question asks
We need the total gas volume after complete reaction, accounting for what is consumed and what remains (including excess reactant and product).
Step 2: Eliminate $110.0 \text{ cm}^3$
This would imply no gas is consumed at all — clearly incorrect since the reaction converts $SO_{2}$ and $O_{2}$ into $SO_{3}$ , and the total moles of gas decrease.
Step 3: Eliminate $60.0 \text{ cm}^3$
This would suggest only the $SO_{3}$ product remains with no excess $O_{2}$ . But $SO_{2}$ is the limiting reactant, so $O_{2}$ is in excess — $20.0 cm^{3}$ of $O_{2}$ remains unreacted.
Step 4: Eliminate $50.0 \text{ cm}^3$
This is simply the initial volume of $O_{2}$ , which does not correspond to any meaningful calculation of remaining gas after the reaction.
Step 5: Select the correct answer
$SO_{3}$ produced $= 60.0 cm^{3}$ ; excess $O_{2} = 20.0 cm^{3}$ . Total $= 80.0 cm^{3}$ . The correct answer is $80.0 cm^{3}$ .
Question 2
The thermite reaction used in welding is: $2 Al (s) + Fe_{2} O_{3} (s) \to Al_{2} O_{3} (s) + 2 Fe (l)$ Using $A_{r}$ values: Al = 27, Fe = 56, O = 16, which expression correctly calculates the atom economy for producing iron ( $Fe$ )?
No clue? Show me the answer
Correct answerCorrect!Incorrect
A $\frac{112}{102 + 112} \times 100$
Step-by-step walkthrough
Choose a solution method
Method #1Approach 1
Step 1: Recall the atom economy formula
$Atom economy = \frac{M _{r} ( desired product )}{\sum M _{r} ( all products )} \times 100$
Step 2: Identify all products and their molar masses
The two products are $Al_{2} O_{3}$ and $2 Fe$ . $M_{r} (Al_{2} O_{3}) = 2 (27) + 3 (16) = 54 + 48 = 102$ . $M_{r} (2 Fe) = 2 \times 56 = 112$ .
Step 3: Identify the desired product
The desired product is iron (Fe). The coefficient is 2 in the balanced equation, so the relevant molar mass is $2 \times 56 = 112 g mol^{- 1}$ .
Step 4: Write the correct expression
$Atom economy = \frac{112}{102 + 112} \times 100 = \frac{112}{214} \times 100 \approx 52.3%$
Method #2Approach 2
Step 1: Identify what is needed
Atom economy uses the sum of molar masses of ALL products in the denominator, and the molar mass of the desired product only in the numerator — both calculated using stoichiometric coefficients from the balanced equation.
Step 2: Eliminate $\frac{56}{214} \times 100$
This uses the molar mass of a single Fe atom (56) rather than the 2 mol of Fe shown in the equation. The coefficient must be included, so numerator should be $2 \times 56 = 112$ .
Step 3: Eliminate $\frac{102}{214} \times 100$
This uses $Al_{2} O_{3}$ (the unwanted by-product, $M_{r} = 102$ ) as the desired product in the numerator. The desired product is Fe, not $Al_{2} O_{3}$ .
Step 4: Eliminate $\frac{112}{102} \times 100$
This omits $Al_{2} O_{3}$ from the denominator entirely. The denominator must include all products: $102 + 112 = 214$ , not just one product.
Step 5: Select the correct expression
$\frac{112}{102 + 112} \times 100$ correctly places the molar mass of 2 Fe in the numerator and the sum of molar masses of all products in the denominator.

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Step-by-step walkthrough

Step 1: Identify the mole ratio and limiting reactant

Step 2: Determine which reactant is limiting

Step 3: Calculate volumes consumed and produced

Step 4: Calculate total gas volume remaining

Step 1: Identify what the question asks

Step 2: Eliminate $110.0 \text{ cm}^3$

Step 3: Eliminate $60.0 \text{ cm}^3$

Step 4: Eliminate $50.0 \text{ cm}^3$

Step 5: Select the correct answer

Step-by-step walkthrough

Step 1: Recall the atom economy formula

Step 2: Identify all products and their molar masses

Step 3: Identify the desired product

Step 4: Write the correct expression

Step 1: Identify what is needed

Step 2: Eliminate $\frac{56}{214} \times 100$

Step 3: Eliminate $\frac{102}{214} \times 100$

Step 4: Eliminate $\frac{112}{102} \times 100$

Step 5: Select the correct expression

14 more questions in this topic

Step-by-step walkthrough

Step 1: Identify the mole ratio and limiting reactant

Step 2: Determine which reactant is limiting

Step 3: Calculate volumes consumed and produced

Step 4: Calculate total gas volume remaining

Step 1: Identify what the question asks

Step 2: Eliminate $110.0 \text{ cm}^3$

Step 3: Eliminate $60.0 \text{ cm}^3$

Step 4: Eliminate $50.0 \text{ cm}^3$

Step 5: Select the correct answer

Step-by-step walkthrough

Step 1: Recall the atom economy formula

Step 2: Identify all products and their molar masses

Step 3: Identify the desired product

Step 4: Write the correct expression

Step 1: Identify what is needed

Step 2: Eliminate $\frac{56}{214} \times 100$

Step 3: Eliminate $\frac{102}{214} \times 100$

Step 4: Eliminate $\frac{112}{102} \times 100$

Step 5: Select the correct expression

14 more questions in this topic