Question

Calculate the mass of water produced when 9.38 g

g of butane reacts with excess oxygen.

Answer 1

$2C _4 H _{10} ( g ) + 13 O _2 ( g ) → 4 C O _2 ( g ) + 5 H _2 O ( l )$

Moles of butane $=\frac{9.38} {58.12}=0.16 moles$

Moles of oxygen $=\frac{13×0.16}{2}=1.04moles$

Moles of water $=\frac{5×0.16}{2}=0.4moles$

Mass of water $=18.02×0.4=7.208grams$