Calculate the mass and volume of H2 released during the interaction of 4g Ca with HCl?

Answer: 0.2 g, 2.24 l.
Explanation of the solution to the problem: in order to solve this problem, it is first necessary to draw up the equation of the chemical reaction: Ca + 2HCl => CaCl2 + H2.
Then you need to find the amount of calcium substance, for this you need to divide its mass by its molar mass:
n (Ca) = m / M = 4/40 = 0.1 mol.
Now you need to find the amount of hydrogen, it is equal to the amount of calcium, because the reaction shows that they interact in a 1: 1 ratio:
n (H2) = n (Ca) = 0.1 mol.
The last step is to find the mass of hydrogen, for this we multiply the found amount of the substance by its molar mass and volume, multiplying the amount of the substance by the molar volume:
m (H2) = n (H2) * M = 0.1 * 2 = 0.2 g.
V (H2) = n (H2) * Vm = 0.1 * 22.4 = 2.24 liters.




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