Question

A. Calculate the mass of water produced when 7.06 g of butane reacts with excess oxygen.

B.Calculate the mass of butane needed to produce 64.9 g of carbon dioxide.

Answer 1

Balanced chemical equation is;

2 C₄H₁₀ + 13 O₂ -------------> 8 CO₂ + 10 H₂O

A)

no. of moles of butane = 7.06 / 58.124 = 0.12146 mole

since O2 is excess, limiting reagent is butane

from the equation, 2 moles of butane will reacted to produce water = 10 moles

therefore, 0.12146 moles of butane produced water = (0.12146 x 5) = 0.6073 moles

Weight of H2O formed = 0.6073 x 18.015 = 10.941 grams

B)

no. of moles of CO2 = 64.9/44 = 1.475 moles

no. of moles of butane required = 1.475 x (2/8) = 0.36785 moles

mass of butane required = 0.36785 x 58.124 = 21.43 grams