What volume of oxygen gas, in milliliters, is required to react with 0.640 g of SO2 gas at STP? 11.2 mL 22.4 mL 112 mL 224 m

Question

asked Oct 13, 2018 1.0k views

2 Answers

David Myers · Answer 1 · 2018-10-17T11:15:15+0000

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
.

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

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
(0.01)/(2)=5* 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* 10^(-3) mole of oxygen gas contains
22400ml* (5* 10^(-3))=112ml volume of oxygen gas

Therefore, the volume of oxygen gas is, 112 ml