Question

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

A. calculate the mass of water produced when 6.30g of butane (C4H10) reacts with excess oxygen.

B. calculate the mass of butane needed to produce 47.2g of carbon dioxide

Answer 1

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

1 ) from balence chemical reaction 2 mole of C4H10 produce 10 mole of H2O in excess amount ofoxygen

molar mass of C4H10 = 58 g/mol

number of moles of C4H10 = (6.30 g / 58 g/mol) = 0.1086 moles

ratio is 2:10 butane to H2O

moles of H2O = (0.1086 *10) / 2 = 0.543 moles of H2O formed

mass of H2O = 18 * 0.543 = 9.774 g

2) for butane to CO2 ratio is 2:8

moles of CO2 = (47.2g / 44g/mol )= 1.0727 moles of CO2

moles of butane = (1.0727 * 2) / 8 = 0.268 moles

mass of butane = 0.268 * 58 = 15.55g