Question

What volume of oxygen gas, in milliliters, is required to react with 0.640 g of SO2 gas at STP?

Answer 1

Answer : The correct option is, 112 ml

Solution : Given,

Mass of $SO_2$ = 0.640 g

Molar mass of $SO_2$ = 64 g/mole

First we have to calculate the moles of $SO_2$.

$\text{Moles of }SO_2=\frac{\text{Mass of }SO_2}{\text{Molar mass of }SO_2}=\frac{0.640g}{64g/mole}=0.01moles$

Now we have to calculate the moles of oxygen gas.

The balanced chemical reaction will be,

$2SO_2+O_2\rightarrow 2SO_3$

From the reaction, we conclude that

As, 2 moles of $SO_2$ react with 1 mole of $O_2$ gas

So, 0.01 moles of $SO_2$ react with $\frac{0.01}{2}=5\times 10^{-3}$ mole of $O_2$ gas

Now we have to calculate the volume of oxygen gas.

As, 1 mole of oxygen gas contains 22400 ml volume of oxygen gas

So, $5\times 10^{-3}$ mole of oxygen gas contains $22400ml\times (5\times 10^{-3})=112ml$ volume of oxygen gas

Therefore, the volume of oxygen gas is, 112 ml

Answer 2

SO2 + O2 ---> SO3. Moles of SO2 = 0.64/32+16X2 = 0.01mole. According to stoichiometry, moles of SO2 = moles of O2 = 0.01mol. Therefore volume of oxygen gas at STP = 22.4 x moles of O2 = 22.4 x 0.01 = 0.224 = 224mL.