Question

What is the change in volume when 4 grams of H2 and 32 grams of O2 react to completion at STP.

2H2 (g) + O2 (g) ⟶ 2 H2O(g)

Group of answer choices

The system volume does not change

The system decreases in volume by about 44 liters

The system decreases in volume by about 22 liters

The system increases in volume by about 44 liters

Answer 1

Answer:

The system decreases in volume by about 22 liters

Explanation:

H2 mols= 4/2= 2mol

O2 mols= 32/32=1 mol

if completely reacted,no limiting reactant.

water=2mols

PV=nRT

n/V=P/RT

V at STP=22 l/mol

volume of 3mol=66

volume of 2 mol=44

change=66-44=22

Answer 2

The system decreases in volume by 22 litres (Option C)

Given:

4 grams of $H_{2}$

32 grams of $O_{2}$

react to completion at STP

To find:

Volume change of the system

Solution:

H2 mols of this system is = 4/2 = 2 mols

And O2 mols in this system = 32/32 = 1 mol

and water produced 2 mols

as we know molar volume at STP is 22.4 l/mol

so the volume of 3 mols reactants = 66

and volume of 2 mols H2O = 44

so change in volume is = 66-44 = 22 litres

So the volume of system decreases by 22 litres

#SPJ3