Question

in an experiment, 16.0 g of so₂ is treated with 6.0 g of o₂. if so₃ is produced with a 75.0% yield, what mass of so₃ was produced? choose the closest value.
2 so₂(g) + o₂(g) → 2 so₃(g)
molar masses (in g mol⁻¹) 64.06 32.00 80.06
a. 20.0 g
b. 30.0 g
c. 22.5 g
d. 26.7 g
e. 15.0 g

Explanation:

Step1: Calculate moles of reactants

Moles of $SO_2=\frac{16.0\ g}{64.06\ g/mol}\approx0.25\ mol$
Moles of $O_2 = \frac{6.0\ g}{32.00\ g/mol}= 0.1875\ mol$

Step2: Determine limiting reactant

From the balanced equation $2SO_2(g)+O_2(g)
ightarrow2SO_3(g)$, the mole - ratio of $SO_2$ to $O_2$ is 2:1.
For 0.25 mol of $SO_2$, moles of $O_2$ required $=\frac{0.25\ mol}{2}=0.125\ mol$. Since $0.125\ mol<0.1875\ mol$, $SO_2$ is the limiting reactant.

Step3: Calculate theoretical yield of $SO_3$

From the balanced equation, moles of $SO_3$ produced from 0.25 mol of $SO_2$ (theoretical) is 0.25 mol (1:1 mole - ratio of $SO_2$ to $SO_3$).
Theoretical mass of $SO_3 = 0.25\ mol\times80.06\ g/mol = 20.015\ g$

Step4: Calculate actual yield of $SO_3$

Actual yield = Theoretical yield $\times$ Percent yield
Actual yield $=20.015\ g\times0.75 = 15.01125\ g\approx15.0\ g$

Answer:

E. 15.0 g