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

Question

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

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Thiên Di 5 months 2021-08-15T06:28:21+00:00 1 Answers 5 views 0

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    2021-08-15T06:29:46+00:00

    Answer:

    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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Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )