Question

For each of the following precipitation reactions, calculate how many grams of the first reactant are necessary to completely react with 17.0 g of the second reactant. Part A 2KI(aq)+Pb(NO3)2(aq)→PbI2(s)+2KNO3(aq) Part B Na2CO3(aq)+CuCl2(aq)→CuCO3(s)+2NaCl(aq) Part C K2SO4(aq)+Sr(NO3)2(aq)→SrSO4(s)+2KNO3(aq)

Answer 1

a)

Molar mass of Pb(NO3)2 = 1*MM(Pb) + 2*MM(N) + 6*MM(O)

= 1*207.2 + 2*14.01 + 6*16.0

= 331.22 g/mol

mass of Pb(NO3)2 = 17 g

mol of Pb(NO3)2 = (mass)/(molar mass)

= 17/331.22

= 0.0513 mol

From balanced chemical reaction, we see that

when 1 mol of Pb(NO3)2 reacts, 2 mol of KI is reacting

mol of KI reacts = (2/1)* moles of Pb(NO3)2

= (2/1)*0.0513

= 0.1027 mol

Molar mass of KI = 1*MM(K) + 1*MM(I)

= 1*39.1 + 1*126.9

= 166 g/mol

mass of KI = number of mol * molar mass

= 0.1027*166

= 17.04 g

Answer: 17.0 g

b)

Molar mass of CuCl2 = 1*MM(Cu) + 2*MM(Cl)

= 1*63.55 + 2*35.45

= 134.45 g/mol

mass of CuCl2 = 17 g

mol of CuCl2 = (mass)/(molar mass)

= 17/134.45

= 0.1264 mol

From balanced chemical reaction, we see that

when 1 mol of CuCl2 reacts, 1 mol of Na2CO3 is reacting

mol of Na2CO3 reacts = moles of CuCl2

= 0.1264 mol

Molar mass of Na2CO3 = 2*MM(Na) + 1*MM(C) + 3*MM(O)

= 2*22.99 + 1*12.01 + 3*16.0

= 105.99 g/mol

mass of Na2CO3 = number of mol * molar mass

= 0.1264*105.99

= 13.4015 g

Answer: 13.4 g

c)

Molar mass of Sr(NO3)2 = 1*MM(Sr) + 2*MM(N) + 6*MM(O)

= 1*87.62 + 2*14.01 + 6*16.0

= 211.64 g/mol

mass of Sr(NO3)2 = 17 g

mol of Sr(NO3)2 = (mass)/(molar mass)

= 17/211.64

= 0.0803 mol

From balanced chemical reaction, we see that

when 1 mol of Sr(NO3)2 reacts, 1 mol of K2SO4 is reacting

mol of K2SO4 reacts = moles of Sr(NO3)2

= 0.0803 mol

Molar mass of K2SO4 = 2*MM(K) + 1*MM(S) + 4*MM(O)

= 2*39.1 + 1*32.07 + 4*16.0

= 174.27 g/mol

mass of K2SO4 = number of mol * molar mass

= 0.0803*174.27

= 14.0 g

Answer: 14.0 g