Question

For each of the reactions, calculate the mass (in grams) of the product formed when 15.34 g of the bolded reactant completely reacts. Assume that there is more than enough of the other reactant.

1. 2K(s)+Cl2(g)−−−−−→2KCl(s)

2. 2K(s)+Br2(l)−−−−−→2KBr(s)

3. 4Cr(s)+3O2(g)−−−−−→2Cr2O3(s)

4. 2Sr(s)−−−−+O2(g)→2SrO(s)

Answer 1

1)

Molar mass of Cl2 = 70.9 g/mol

mass of Cl2 = 15.34 g

mol of Cl2 = (mass)/(molar mass)

= 15.34/70.9

= 0.2164 mol

According to balanced equation

mol of KCl formed = (2/1)* moles of Cl2

= (2/1)*0.2164

= 0.4327 mol

Molar mass of KCl,

MM = 1*MM(K) + 1*MM(Cl)

= 1*39.1 + 1*35.45

= 74.55 g/mol

mass of KCl = number of mol * molar mass

= 0.4327*74.55

= 32.26 g

Answer: 32.26 g

2)

Molar mass of Br2 = 159.8 g/mol

mass of Br2 = 15.34 g

mol of Br2 = (mass)/(molar mass)

= 15.34/1.598*10^2

= 9.599*10^-2 mol

According to balanced equation

mol of KBr formed = (2/1)* moles of Br2

= (2/1)*9.599*10^-2

= 0.192 mol

Molar mass of KBr,

MM = 1*MM(K) + 1*MM(Br)

= 1*39.1 + 1*79.9

= 119 g/mol

mass of KBr = number of mol * molar mass

= 0.192*1.19*10^2

= 22.85 g

Answer: 22.85 g

3)

Molar mass of O2 = 32 g/mol

mass of O2 = 15.34 g

mol of O2 = (mass)/(molar mass)

= 15.34/32

= 0.4794 mol

According to balanced equation

mol of Cr2O3 formed = (2/3)* moles of O2

= (2/3)*0.4794

= 0.3196 mol

Molar mass of Cr2O3,

MM = 2*MM(Cr) + 3*MM(O)

= 2*52.0 + 3*16.0

= 152 g/mol

mass of Cr2O3 = number of mol * molar mass

= 0.3196*1.52*10^2

= 48.58 g

Answer: 48.58 g

4)

Molar mass of Sr = 87.62 g/mol

mass of Sr = 15.34 g

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

= 15.34/87.62

= 0.1751 mol

According to balanced equation

mol of SrO formed = moles of Sr

= 0.1751 mol

Molar mass of SrO,

MM = 1*MM(Sr) + 1*MM(O)

= 1*87.62 + 1*16.0

= 103.62 g/mol

mass of SrO = number of mol * molar mass

= 0.1751*1.036*10^2

= 18.14 g

Answer: 18.14 g