Question

What volume of product is produced when 20.0 Mg of liquid hydrogen reacts with

excess liquid oxygen in a rocket propellant reaction? The gas is formed at 101.325 kPa and 40˚ C

Answer 1

Answer: The volume of product formed is 0.26 L

Step-by-step explanation:

`\text{Moles of solute}=\frac{\text{given mass}}*{\text{Molar Mass}}`

`\text{Moles of} H_2=(0.02g)/(2g/mol)=0.01moles`

`2H_2(l)+O_2(l)\rightarrow 2H_2O(g)`

As
`O_2` is the the excess reagent,
`H_2` is the limiting reagent as it limits the formation of product.

According to stoichiometry :

2 moles of
`H_2` give = 2 moles of
`H_2O`

Thus 0.01 moles of
`H_2` will give =
`(2)/(2)* 0.01=0.01moles` of
`H_2O`

According to ideal gas equation:

`PV=nRT`

P = pressure of gas = 101.325 kPa = 1 atm

V = Volume of gas = ?

n = number of moles = 0.01

R = gas constant =
`0.0821Latm/Kmol`

T =temperature =
`40^0C=(40+273)K=313K`

`V=(nRT)/(P)`

`V=(0.01mol* 0.0821L atm/K mol* 313K)/(1atm)=0.26L`

Thus the volume of product formed is 0.26 L