Question

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.

Answer 1

Answer:

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) $2C_4H_{10}(g)+13O_2(g)\rightarrow 10H_2O(g)+8CO_2(g)$

Moles of butane = $\frac{1.77 g}{58 g/mol}=0.0305 mol$

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

Then 0.0305 moles of butane will produce:

$\frac{10}{2}\times 0.0305 mol=0.1525 mol$ 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 = $\frac{71.6 g}{44 g/mol}=1.627 mol$

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:

$\frac{2}{8}\times 1.627 mol=0.4067 mol$ 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.

Answer 2

The balanced chemical reaction is:

2C4H10(g)+13O2(g)->10H2O(g)+8CO2(g) 

Calculate the mass of water produced when 1.77 grams of butane reacts with excessive oxygen?
1.77 g C4H10 (1 mol C4H10/58.14 g C4H10) (10 mol H2O / 2 mol C4H10) ( 1.01 g H2O / 1 mol H2O ) = 0.15 g H2O

Calculate the mass of butane needed to produce 71.6 of carbon dioxide.
71.6 g CO2 (1 mol CO2/ 44.01 g CO2) ( 2 mol C4H10 / 8 mol CO2 ) (58.14 g C4H10 / 1 mol C4H10 ) = 23.65 g C4H10