How many moles of sodium chloride are formed by the interaction of sodium sulfide weighing 12.6 g with hydrochloric acid?

Let’s implement the solution:
1. Let us write the equation according to the problem statement:
m = 12.6 g. Y -?
Na2S + 2HCl = 2NaCl + H2S – ion exchange, hydrogen sulfide is released;
2. Let’s make the calculations:
M (Na2S) = 77.8 g / mol;
M (NaCl) = 58.4 g / mol.
3. Let’s calculate the number of moles of sodium sulfide, if the mass is known:
Y (Na2S) = m / M = 12.6 / 77.8 = 0.16 mol.
4. We make the proportion:
0.16 mol (Na2S) – X mol (NaCl);
-1 mol -2 mol hence, X mol (NaCl) = 0.16 * 2/1 = 0.32 mol.
Answer: the reaction product, sodium chloride, was obtained in an amount of 0.32 mol.

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