Butane, c4h10, reacts with oxygen, o2, to form water, h2o, and carbon dioxide, co2, as shown in the following chemical equation: 2c4h10(g)+13o2(g)-> 10h2o(g)+8co2(g)

calculate the mass of water produced when 1.77 grams of butane reacts with excessive oxygen?
calculate the mass of butane needed to produce 71.6 of carbon dioxide.

2.745 gram of water produced when 1.77 grams of butane reacts with excessive oxygen.

23.58 gram of butane needed to produce 71.6 of carbon dioxide.

Explanation:

1)

Moles of butane =

According to reaction, 2 moles of butane produces 10 moles of water.

Then 0.0305 moles of butane will produce:

of water

Mass of 0.1525 mol of water = 0.1525 mol × 18 g/mol = 2.745 g

2.745 gram of water produced when 1.77 grams of butane reacts with excessive oxygen.

2)Moles of carbon dioxide =

According to reaction 8 moles of carbon dioxide are produced from 2 moles of butane.

Then 1.627 mol of carbon-dioxide will be produced from:

of butane

Mass of 0.4067 moles of butane =  0.4067 mol × 58 g/mol = 23.58 g

23.58 gram of butane needed to produce 71.6 of carbon dioxide.