Question

A student needs 62.5 g of zinc sulfide, a white pigment, for an art project. He can synthesize it using the reaction: Na2S (aq) + Zn(NO3)2 (aq) = ZnS(s) + NaNO3 (aq) (not balanced) (78.04) (189.4) (97.44) (85.00) How many grams of zinc nitrate will he need if the percent yield of zinc sulfide is 85.0 %?

Answer 1

The balanced equation is   Na₂S (aq) + Zn(NO₃)₂ (aq) ZnS(s) + 2NaNO₃ (aq)

Molar mass of Zn(NO₃)₂ = At.mass of Zn+ (2xAt.mass of N ) + (6xAt.mass of O )

                                    = 65.4 +(2x14) + (6x16)

                                   = 189.4 g/mol

Molar mass of ZnS = At.mass of Zn + At.mass of S

                            = 65.4 + 32

                            = 97.4 g/mol

Given that the mass of ZnS is 62.5 g

This is the mass for 85.0 % yield.

For 100 % yield the mass of ZnS is =(62.5 x100 ) / 85.0

                                                   = 73.53 g

According to the balanced equation,

1 mole of Zn(NO₃)₂ (aq) produces 1 mole of ZnS(s)

                           OR

1x189.4 g of Zn(NO₃)₂ (aq) produces 1x97.4 g of ZnS(s)

Y g of Zn(NO₃)₂ (aq) produces 73.53 g of ZnS(s)

Y = (73.53x1x189.4) / (1x97.4)

   = 143.0 g

Therefore the mass of zinc nitrate neede is 143.0 g