Question

What volume of butane can be produced from the reaction of 13.45 g of carbon and 17.65 l of hydrogen gas

Answer 1

Complete Question

What volume of butane can be produced from the reaction of 13.45 g of carbon and 17.65 L of hydrogen gas at STP ?

Answer:

The volume of butane produced is
`V_b = 3.54 \ L`

Step-by-step explanation:

From the question we are told that

The mass of carbon is
`m_c = 13.45 \ g`

The volume of hydrogen is
`V_h = 17.65\ L`

The number of moles of carbon is mathematically evaluated as

`n_c = (m_c)/(M_c)`

Where
`M_c` is the molar mass of carbon which is a constant with value
`M_c = 12\ g/mol`

So

`n_c = (13.45)/(12)`

`n_c = 1.12`

Now since the volume of the hydrogen is measured at standard temperature and pressure

Hence the number of moles of hydrogen is

`n_h = (V_h )/(22.4)`

Where 22.4 is a constant

`n_h = (17.65)/(22.4)`

`n_h = 0.79`

Comparing
`n_h \ and\ n_c` we see that hydrogen is the limiting reactant

because
`n_h` is greater than
`n_c`

The chemical equation for this reaction is

`4C_((s)) + 5H_2-{(g)} ----> C_5 H_10_((g))`

looking at chemical equation we see that

5 moles of hydrogen gas reacts with 4 moles of carbon to produce 1 mole of butane

This implies that

0.79 moles react with 1.12 moles of carbon to produce x moles of butane

Therefore

`x = (0.79)/(5)`

`x = 0.158 \ moles`

Now since this reaction is carried out at standard temperature and pressure the volume of butane produced will be

`V_b = 0.158 * 22.4`

`V_b = 3.54 \ L`