Question

The substances butane (C4H10) and oxygen gas react to form carbon dioxide and water.

Unbalanced equation: C4H10 (g) + O2 (g) CO2 (g) + H2O (g)

In one reaction, 48.0 g of H2O is produced. What amount (in mol) of O2 was consumed? What mass (in grams) of CO2 is produced? mol O2 consumed g CO2 produced

Answer 1

Answer:

Given reaction is

C4H10(g) + O2(g) -------> CO(g) + H2O(g)

The balanced equation is

2C4H10 (g) + 13O2 (g) -------> 8CO2 (g) + 10H2O (g)

Given mass of H2O produced=48 g and

molar mass of H2O=18 g/mol.

Moles of H2O produced=mass/molar mass=48 g/18 g/mol=2.667 mol.

From balanced equation, moles of O2 consumed=(13/10) mol H2O produced=(13/10) x 2.667 mol=3.4667 mol O2 consumed.

Moles of CO2 produced=(8/10) moles of H2O produced=(8/10) x 2.667 mol=2.1333 mol.

Molar mass of CO2=44.0095 g/mol.

Therefore mass of CO2 produced=moles x molar mass=2.133 mol x 44.0095 g/mol=93.887 g CO2 produced.

Mol O2 consumed=3.467 mol.

Mass of CO2 produced=93.887 g.

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