What volume of hydrogen will be released when potassium interacts with sulfate acid with an amount of 5 mol of a substance?
1. Let’s write the equation of interaction of potassium with sulfuric acid:
2K + H2SO4 = K2SO4 + H2 ↑;
2.In the reaction equation, there are unit coefficients in front of the acid and the evolving hydrogen, so the amount of hydrogen is:
n (H2) = n (H2SO4) = 5 mol;
3.Calculate the volume of hydrogen:
V (H2) = n (H2) * Vm = 5 * 22.4 = 112 HP
Answer: 112 liters.
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