Question

Determine the volume of SO2 (at STP) formed from the reaction of 96.7 g of FeS2 and 55.0 L of O2 (at 398 K and 1.20 atm). The molar mass of FeS2 is 119.99 g/mol. 4 FeS2(s) + 11 O2(g) → 2 Fe2O3(s) + 8 SO2(g)

Answer 1

Answer:  32.9 Liters

Explanation:

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}$

1. moles of $FeS_2=\frac{96.7g}{119.99g/mol}=0.806mol$

2. moles of $O_2$

$PV=nRT$

P = pressure of the gas = 1.20  atm

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature of the gas = $398K$

$1.20\times 55.0=n\times 0.0821\times 398$

$n=2.02$

$4FeS_2(s)+11O_2(g)\rightarrow 2Fe_2O_3(s)+8SO_2(g)$

According to stoichiometry:

11 moles of oxygen reacts with 4 moles of $FeS_2$

Thus 2.02 moles of oxygen reacts with =$\frac{4}{11}\times 2.02=0.73$ moles of $FeS_2$

Thus oxygen acts as limiting reagent and $FeS_2$ is excess reagent.

As 11 moles of oxygen gives = 8 moles of $SO_2$

2.02 moles of oxygen gives =$\frac{8}{11}\times 2.02=1.47$ moles of $SO_2$

$PV=nRT$

P = pressure of the gas = 1  atm (at STP)

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature of the gas = $273K$   (at STP)

$1\times V=1.47\times 0.0821\times 273$

$V=32.9L$

Thus volume of $SO_2$ (at STP) formed from the reaction is 32.9 L