Question

Сколько г сульфида цинка образуется при взаимодействии 15,6 г сульфида натрия с избытком хлорида цинка

Answer 1

Реакция:
Na2S + ZnCl2 = 2 NaCl + ZnS↓
Из уравнения реакции следует, что n(Na2S) = n(ZnS).
Определим кол-во вещества сульфида цинка, найдя его для сульфида натрия:
n(ZnS) = n(Na2S) = m(Na2S)/M(Na2S) = 15,6/(23*2+32)=0,2(моль).
Зная количество вещества сульфида цинка, найдём его массу:
m(ZnS) = n(ZnS)*M(ZnS) = 0,2*(65+32) = 19,4(г).
Ответ: m(ZnS) = 19,4 г.

Answer 2

Calculation of Zinc Sulfide Formation

To calculate the amount of zinc sulfide (ZnS) formed when 15.6 g of sodium sulfide (Na2S) reacts with an excess of zinc chloride (ZnCl2), we can use the stoichiometry of the reaction between sodium sulfide and zinc chloride.

The balanced chemical equation for the reaction is: Na2S + ZnCl2 → ZnS + 2NaCl

According to the stoichiometry of the reaction, 1 mole of Na2S reacts with 1 mole of ZnCl2 to produce 1 mole of ZnS.

First, we need to find the number of moles of Na2S: Number of moles = Mass / Molar mass

The molar mass of Na2S is 78.045 g/mol.

Number of moles of Na2S = 15.6 g / 78.045 g/mol = 0.2 moles

Since the reaction between Na2S and ZnCl2 occurs in a 1:1 ratio, the number of moles of ZnS formed will be the same as the number of moles of Na2S.

Therefore, 0.2 moles of ZnS will be formed when 15.6 g of Na2S reacts with an excess of ZnCl2.

Conclusion

When 15.6 g of sodium sulfide reacts with an excess of zinc chloride, 0.2 moles of zinc sulfide (ZnS) are formed.