Calculate the mass of water produced when 1.41 g
of butane reacts with excess oxygen.
2C4H10+13O2→8CO2+10H2O2C_4H_{10}+13O_2 \to 8CO_2+10H_2O2C4H10+13O2→8CO2+10H2O
Molar mass of butane =2×58=116=2\times 58=116=2×58=116
number of moles of butane =1.41/116=0.0122=1.41/116=0.0122=1.41/116=0.0122
molar mass of water=10×18=180=10\times 18=180=10×18=180
mole ratio 2C4H10:10H2O=2:10=1:52C_4H_{10}:10H_2O=2:10=1:52C4H10:10H2O=2:10=1:5
moles of water =5×0.0122=0.061=5\times 0.0122=0.061=5×0.0122=0.061
mass of water produced =0.061×180=10.98g=0.061\times 180=10.98g=0.061×180=10.98g
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