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.

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

Tringa0 Tringa0 · Answer 2 · 2025-03-26T23:00:56-04:00

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) [tex]2C_4H_{10}(g)+13O_2(g)\rightarrow 10H_2O(g)+8CO_2(g)[/tex]

Moles of butane = [tex]\frac{1.77 g}{58 g/mol}=0.0305 mol[/tex]

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

Then 0.0305 moles of butane will produce:

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

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:

[tex]\frac{2}{8}\times 1.627 mol=0.4067 mol[/tex] 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.