Question

Zinc reacts with hydrochloric acid according to the following equation: Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g) In a certain experiment, a 3.77 g sample of impure zinc is reacted with excess hydrochloric acid. If 1.09 L of H2 gas is collected at STP, what percentage of the original sample was zinc? Assume that the impurities do not react with HCl.

Answer 1

approximately 69.24% of the original sample was zinc.

Step-by-step explanation:

According to the equation, 1 mole of zinc reacts with 2 moles of HCl to produce 1 mole of ZnCl2 and 1 mole of H2.

We can use the Ideal Gas Law PV = nRT to find out the number of moles of H2 produced. Given that the pressure P = 1atm and temperature T = 273K (STP).

n = PV / RT = (1.09L) * (1atm) / (8.314J/mol * 273K) = 0.04 moles

Now we know that 0.04 moles of H2 is produced, we can use the balanced equation to find out the number of moles of zinc that reacted.

0.04 moles H2 is produced for every 1 mole of Zn.

thus 0.04 moles of zinc reacted.

The mass of zinc that reacted can be calculated by multiplying the number of moles by the molar mass of zinc (65.38g/mol)

0.04 moles * 65.38 g/mol = 2.6152 g

Now we can calculate the percentage of zinc in the original sample using the following equation:

% zinc = (mass of zinc reacted / mass of original sample) * 100

= (2.6152g / 3.77g) * 100 = 69.24%

So approximately 69.24% of the original sample was zinc.