Question

2C4H10(g)+13O2(g)→10H2O(g)+8CO2(g); Molar Mass of Butane 58.12g/mol

Calculate the mass of butane needed to produce 90.9 g of carbon dioxide.

Calculate the mass of butane needed to produce 90.9 g of carbon dioxide.

Answer 1

Answer : The mass of Butane needed is 29.946 g.

Solution : Given,

Molar mass of butane = 58.12 g/mole

Molar mass of carbon dioxide = 44.01 g/mole

Mass of carbon dioxide = 90.9 g

The Given net balanced chemical reaction is,

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

First we have to calculate the moles of carbon dioxide.

`\text{ Moles of }CO_2=\frac{\text{ Given mass of }CO_2}{\text{ Molar mass of }CO_2 }`

`\text{ Moles of }CO_2=(90.9g)/(44.01g/mole)=2.061moles`

From the given chemical reaction, we conclude that

8 moles of
`CO_2` produced from 2 moles
`C_4H_(10)`

2.061 moles of
`CO_2` produced from
`(2moles)/(8moles)* 2.061moles=0.51525moles` of
`C_4H_(10)`

The moles of
`C_4H_(10)` = 0.51525 moles

Now we have to calculate the mass of
`C_4H_(10)` needed.

Mass of Butane = Moles of Butane × Molar mass of Butane

Mass of Butane = 0.51525 moles × 58.12 g/mole = 29.946 g

Therefore, the mass of Butane needed is 29.946 g.