2.14g of sodium sulfide react with excess hydrochloric acid to form sodium chloride and dihydrogen sulfide. BALANCE IT TO AN EQUATION

2.14 Na2S + 1.99 HCl > 3.20 NaCl + 0.935 H2S

Explanation:

The Balanced equation is

Na2S + 2HCl = 2NaCl + H2S

As we can see -

78.0452 g/mol of sodium sulfide reacts with 2 mole of HCl.

Mass of one mole of HCl = 36.458 g/mol

So, 78.0452 g/mol of sodium sulfide reacts with 2* 36.458 g/mol of HCL to produce 2* 58.44 g of NaCl and 34.1 g/mol of H2S

Thus,

2.14g of sodium sulfide will react with (2* 36.458/78.0452)* 2.14 of HCl

2.14g of sodium sulfide will react with 1.99 g of HCl

2.14 Na2S + 1.99 HCl > (2* 58.44/78.0452) *2.14 NaCl + (34.1/78.0452) * 2.14 H2S

2.14 Na2S + 1.99 HCl > 3.20 NaCl + 0.935 H2S

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explanation: