Question

limiting reactants da

1. given the reaction: $ce{s + o_{2} -> so_{2}}$

a) how many grams of $ce{so_{2}}$ can be produced from 109.98 grams of sulfur(s)?
b) how many grams of $ce{so_{2}}$ can be produced from 86.39 grams of oxygen ($ce{o_{2}}$)?
c) which element is the limiting reactant (circle one) sulfur or oxygen
d) how much product would be produced in the above reaction?

Explanation:

Step1: Molar mass of S, O₂, SO₂

Molar mass of S: $32.07\ \text{g/mol}$; Molar mass of $\text{O}_2$: $2\times16.00=32.00\ \text{g/mol}$; Molar mass of $\text{SO}_2$: $32.07 + 2\times16.00=64.07\ \text{g/mol}$

Step2: Solve part (a): Calculate SO₂ from S

Moles of S: $\frac{109.98\ \text{g}}{32.07\ \text{g/mol}}=3.43\ \text{mol}$
Mole ratio S:SO₂ = 1:1, so moles of $\text{SO}_2=3.43\ \text{mol}$
Mass of $\text{SO}_2$: $3.43\ \text{mol} \times 64.07\ \text{g/mol}=219.76\ \text{g}$

Step3: Solve part (b): Calculate SO₂ from O₂

Moles of $\text{O}_2$: $\frac{86.39\ \text{g}}{32.00\ \text{g/mol}}=2.70\ \text{mol}$
Mole ratio $\text{O}_2$:$\text{SO}_2$ = 1:1, so moles of $\text{SO}_2=2.70\ \text{mol}$
Mass of $\text{SO}_2$: $2.70\ \text{mol} \times 64.07\ \text{g/mol}=172.99\ \text{g}$

Step4: Identify limiting reactant (part c)

Oxygen produces less $\text{SO}_2$, so it is limiting.

Step5: Determine actual product (part d)

The limiting reactant dictates product yield, so use the smaller value from (a)/(b).

Answer:

a) $219.76\ \text{g}$
b) $173.0\ \text{g}$
c) Oxygen
d) $173.0\ \text{g}$