Question

1.) When 3.31 g of AuCN react with 0.55 g of Zn, 2.31 g of pure gold (Au) are collected. Determine the limiting reagent, theoretical yield of gold, and the percent yield for the reaction.

2AuCN(aq) + Zn(s) —> 2Au(s) + Zn(CN)2(aq)

Limiting reactant ____________ Theoretical yield ___________ Percent yield __________

2.) What mass of Ca(OH)2, in grams, is required to prepare 100.0 mL of 2.25 M Ca(OH)2?

Answer 1

1)

a)

Molar mass of AuCN = 1*MM(Au) + 1*MM(C) + 1*MM(N)

= 1*197.0 + 1*12.01 + 1*14.01

= 223.02 g/mol

mass of AuCN = 3.31 g

we have below equation to be used:

number of mol of AuCN,

n = mass of AuCN/molar mass of AuCN

=(3.31 g)/(223.02 g/mol)

= 1.484*10^-2 mol

Molar mass of Zn = 65.38 g/mol

mass of Zn = 0.55 g

we have below equation to be used:

number of mol of Zn,

n = mass of Zn/molar mass of Zn

=(0.55 g)/(65.38 g/mol)

= 8.412*10^-3 mol

we have the Balanced chemical equation as:

2 AuCN + Zn ---> 2 Au + Zn(CN)2

2 mol of AuCN reacts with 1 mol of Zn

for 0.0148 mol of AuCN, 0.0074 mol of Zn is required

But we have 0.0084 mol of Zn

so, AuCN is limiting reagent

Answer: Limiting reactant: AuCN

b)

we will use AuCN in further calculation

Molar mass of Au = 197 g/mol

From balanced chemical reaction, we see that

when 2 mol of AuCN reacts, 2 mol of Au is formed

mol of Au formed = (2/2)* moles of AuCN

= (2/2)*0.0148

= 1.484*10^-2 mol

we have below equation to be used:

mass of Au = number of mol * molar mass

= 1.484*10^-2*1.97*10^2

= 2.924 g

Answer: theoretical yield = 2.92 g

c)

% yield = actual mass*100/theoretical mass

= 2.31*100/2.924

= 79.0 %

Answer: percent yield = 79.0 %

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