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

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

Answer 2

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.

Step-by-step explanation:

1)
`2C_4H_(10)(g)+13O_2(g)\rightarrow 10H_2O(g)+8CO_2(g)`

Moles of butane =
`(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:

`(10)/(2)* 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 =
`(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:

`(2)/(8)* 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.