Question

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

Answer 1

Answer: The volume of product formed is 0.26 L

Explanation:

$\text{Moles of solute}=\frac{\text{given mass}}\times{\text{Molar Mass}}$    

$\text{Moles of} H_2=\frac{0.02g}{2g/mol}=0.01moles$

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

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 =$\frac{2}{2}\times 0.01=0.01moles$  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 =$40^0C=(40+273)K=313K$

$V=\frac{nRT}{P}$

$V=\frac{0.01mol\times 0.0821L atm/K mol\times 313K}{1atm}=0.26L$

Thus the volume of product formed is 0.26 L