Answer to Question #248504 in General Chemistry for bDC

Question

Calculate the mass of water produced when 1.41 g

of butane reacts with excess oxygen.

Answer 1

$2C_4H_{10}+13O_2 \to 8CO_2+10H_2O$

Molar mass of butane $=2\times 58=116$

number of moles of butane $=1.41/116=0.0122$

molar mass of water$=10\times 18=180$

mole ratio $2C_4H_{10}:10H_2O=2:10=1:5$

moles of water $=5\times 0.0122=0.061$

mass of water produced $=0.061\times 180=10.98g$