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

March 30, 2021 | education

| 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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