Calculate the volume of hydrogen obtained by the interaction of 14.4 zinc containing 10% impurities with an excess of hydrochloric acid.
m (Zn) = 14.4 g
ω approx. = 10%
V (H2) -?
1) Write the reaction equation:
Zn + 2HCl => ZnCl2 + H2 ↑;
2) Find the mass fraction of Zn without impurities:
ω clean. (Zn) = 100% – ω approx. = 100% – 10% = 90%;
3) Find the mass Zn:
m clean (Zn) = ω (Zn) * m (Zn) / 100% = 90% * 14.4 / 100% = 12.96 g;
4) Find the amount of substance Zn:
n (Zn) = m (Zn) / Mr (Zn) = 12.96 / 65 = 0.2 mol;
5) Find the amount of substance H2:
n (H2) = n (Zn) = 0.2 mol;
6) Find the volume H2:
V (H2) = n (H2) * Vm = 0.2 * 22.4 = 4.48 liters.
Answer: The volume of H2 is 4.48 liters.
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