Question

When methane (CH4CH4) burns, it reacts with oxygen gas to produce carbon dioxide and water. The unbalanced equation for this reaction is

CH4(g)+O2(g)→CO2(g)+H2O(g)CH4(g)+O2(g)→CO2(g)+H2O(g)

This type of reaction is referred to as a complete combustion reaction.

Part A

What mass of carbon dioxide is produced from the complete combustion of 1.80×10⁻³ gg of methane?

Part B

What mass of water is produced from the complete combustion of 1.80×10⁻³ gg of methane?

Part C

What mass of oxygen is needed for the complete combustion of 1.80×10⁻³ gg of methane?

Answer 1

the balanced equation is as follows

CH4 + 2 O2 -------------------------> CO2 + 2 H2O

no of moles = mass / molar mass

mass = 1.80*10-3 g = 0.00180 g

molar mass = 16.04 g/mol

for CH4 = 0.0018 / 16.04 => 0.000112219 moles

A) 1 mol CH4 ----------------------> 1 mol CO2

0.000112219 mol --------------->?

=> 0.000112219 * 1 / 1

=> 0.000112219 moles

mass of CO2 => 0.000112219 mol * 44 g/mol => 0.004937 g => 4.94*10^-3 g

B) 1 mol CH4 ----------------------> 2 mol H2O

0.000112219 mol --------------->?

=> 0.000112219 * 2 / 1

=> 0.0002244 moles

mass of H2O => 0.0002244 mol * 18 g/mol => 0.004039 g => 4.04*10^-3 g

C) 1 mol CH4 ----------------------> 2 mol O2

0.000112219 mol --------------->?

=> 0.000112219 * 2 / 1

=> 0.0002244 moles

mass of O2 => 0.0002244 mol * 32 g/mol => 0.007182 g => 7.18*10^-3 g