Question

A 0.2500 g sample of an alloy reacts with to form hydrogen gas: 2Al(s) + 6H+(aq) 2Al3+(aq) + 3H2(g) Zn(s) + 2H+(aq) Zn2+(aq) + H2(g) The hydrogen produced has a volume of 0.147 L at 25ºC and 755 mm Hg. What is the percentage of zinc in the alloy

Answer 1

Answer: The mass percent of zinc in the alloy is 78.68 %

Explanation:

We are given:

Mass of sample of alloy = 0.2500 g

Let the mass of aluminium be 'x' grams and mass of zinc will be (0.2500 - x)  g

To calculate the amount of hydrogen gas produced, we use the equation given by ideal gas which follows:

$PV=nRT$

where,

P = pressure of the gas = 755 mmHg  

V = Volume of the gas = 0.147 L

T = Temperature of the gas = $25^oC=[25+273]K=298K$

R = Gas constant = $62.3637\text{ L.mmHg }mol^{-1}K^{-1}$

n = number of moles of hydrogen gas = ?

Putting values in above equation, we get:

$755mmHg\times 0.147L=n\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 298K\\\\n=\frac{755\times 0.147}{62.3637\times 298}=0.00597mol$

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$    .....(1)

• For Aluminium:

Molar mass of aluminium = 27 g/mol

Putting values in equation 1, we get:

$\text{Moles of aluminium}=\frac{x}{27}mol$

The chemical equation follows:

$2Al(s)+6H^+(aq.)\rightarrow 2Al^{3+}(aq.)+3H_2(g)$

By Stoichiometry of the reaction:

2 moles of aluminium produces 3 moles of hydrogen gas

So, $\frac{x}{27}$ moles of aluminium will produce = $\frac{3}{2}\times \frac{x}{27}=\frac{3x}{54}mol$ of hydrogen gas

• For Zinc:

Molar mass of zinc = 65.4 g/mol

Putting values in equation 1, we get:

$\text{Moles of zinc}=\frac{(0.25-x)}{65.4}mol$

The chemical equation follows:

$Zn(s)+2H^+(aq.)\rightarrow Zn^{2+}(aq.)+H_2(g)$

By Stoichiometry of the reaction:

1 mole of zinc produces 1 moles of hydrogen gas

So, $\frac{(0.25-x)}{65.4}$ moles of zinc will produce = $\frac{1}{1}\times \frac{(0.25-x)}{65.4}=\frac{(0.25-x)}{65.4}mol$ of hydrogen gas

• Equating the moles of hydrogen gas:

$\Rightarrow 0.00597=\frac{3x}{54}+\frac{(0.25-x)}{65.4}\\\\x=0.0533g$

To calculate the mass percentage of zinc in alloy, we use the equation:

$\text{Mass percent of zinc}=\frac{\text{Mass of zinc}}{\text{Mass of alloy}}\times 100$

Mass of zinc = (0.2500 - x) = [0.2500 - 0.0533] = 0.1967 g

Mass of alloy = 0.2500 g

Putting values in above equation, we get:

$\text{Mass percent of zinc in alloy}=\frac{0.1967g}{0.2500g}\times 100=78.68\%$

Hence, the mass percent of zinc in the alloy is 78.68 %