According to the reaction scheme H2S + H2SO4-S + H2O, if hydrogen sulfide with a volume of 3.36 enters into the reaction, then sulfur is formed with a mass
Let’s implement the solution:
1. By the condition of the problem, we write the equation:
V = 3.36 liters. X g -?
3H2S + H2SO4 = 4S + 4H2O – OBP, sulfur was obtained;
M (H2S) = 34 g / mol;
M (S) = 32 g / mol.
3. Let’s make the proportions:
1 mol of gas at normal level – 22.4 liters;
X mol (H2S) – 3.36 L. from here. X mol (H2S) = 1 * 3.36 / 22.4 = 0.15 mol;
0.15 mol (H2S) – X mol (S);
3 mol – 4 mol from here, X mol (S) = 0.15 * 4/3 = 0.2 mol.
4. Find the mass of the product:
m (S) = Y * M = 0.2 * 32 = 6.4 g.
Answer: in the course of the reaction, sulfur with a mass of 6.4 g was obtained.
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