30 October, 06:03

# Identify the limiting reactant when 6.45 g of O2 gas reacts with 1.13 g of H2 gas to produce liquid water.

+4
1. 30 October, 06:15
0
Oxygen is the limiting reactant

Explanation:

2H2 + O2 ⇒ 2H2O

2mol 1mol 2mol

2moles of hydrogen will require 1mol of oxygen to produce 2mol of water

in the question, 6.45g of O2 reacts with 1.13g of H2.

molar mass of O2 = 16*2 = 32g/mol

molar mass of H2 = 1*2=2g/mol

number of moles of O2 = mass/molar mass

= 6.45/32

=0.20156moles

number of moles of H2 = 1.13/2

=0.565moles

but 0.20156moles of O2 will require 0.40312moles of hydrogen

2H2 + O2 ⇒ 2H2O

2mol 1mol 2mol

0.40312mole 0.20156mol

0.40312moles is less than 0.565moles of H2 present, which means hydrogen is in excess hence implying that oxygen is the limiting reactant
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