Question

For each of the reactions, calculate the mass (in grams) of the product formed when 15.30 g of the underlined reactant completely reacts. Assume that there is more than enough of the other reactant

1) 2K(s)+Cl2(g)−−−−−→2KCl(s)

Express your answer using four significant figures

2) 2K(s)+Br2(l)−−−−−→2KBr(s)

Express your answer using four significant figures

3) 4Cr(s)+3O2(g)−−−−−→2Cr2O3(s)

Express your answer using four significant figures

4) 2Sr(s)−−−−+O2(g)→2SrO(s)

Express your answer using four significant figures

Answer 1

A)
2K(s)+Cl2(g) →2KCl(s)

mass of Cl2= 15. 3g
Molar mass of Cl2 = 71 g/mol
Number of moles of Cl2 = mass / molar mass
                                                  = 15.3/ 71
                                                  =0.2155 mol

1 mol of Cl2 forms forms 2 mol of KCl
So,
number of moles of KCl formed = 2* Number of moles of Cl2
                                                                 = 2*0.2155
                                                                 =0. 431 mol
Molar mass of KCl = 74.5 g/mol
mass of KCl = number of moles *molar mass
                         = 0. 431*74.5
                         = 32.11 g
Answer: 32.11 g

b)
2K(s)+Br2(l)−−−−−→2KBr(s)

mass of Br2= 15.3 g
Molar mass of Br2 = 160 g/mol
Number of moles of Br2 = mass / molar mass
                                                  = 15. 3 / 160
                                                  =0.0956 mol

1 mol of Br2 forms forms 2 mol of KBr
So,
number of moles of KBr formed = 2* Number of moles of Br2
                                                                 = 2*0.0956
                                                                 =0.1912 mol
Molar mass of KBr = 119 g/mol
mass of KBr = number of moles *molar mass
                         = 0.1912 *119
                         = 22.75g
Answer: 22.75 g

c)
4Cr(s)+3O2(g)−−−−−→2Cr2O3(s)

mass of O2= 15.3 g
Molar mass of O2 = 32 g/mol
Number of moles of O2 = mass / molar mass
                                                  = 15. 3/ 32
                                                  =0.4781 mol

3 mol of O2 forms forms 2 mol of Cr2O3
So,
number of moles of Cr2O3 formed = (2/3)* Number of moles of O2
                                                                 = (2/3)*0.4781
                                                                 =0.3187 mol
Molar mass of Cr2O3 = 152 g/mol
mass of Cr2O3 = number of moles *molar mass
                         = 0.3187*152
                         = 48.44 g
Answer: 48.44g

d)
2Sr(s)−−−−−+O2(g)→2SrO(s)

mass of Sr= 15.3 g
Molar mass of Sr = 87.62 g/mol
Number of moles of Sr = mass / molar mass
                                                  = 15.3 / 87.62
                                                  =0.1746 mol

2 mol of Sr forms forms 2 mol of SrO
So,
number of moles of SrO formed = 0.1746 mol
Molar mass of SrO = 103.62 g/mol
mass of Cr2O3 = number of moles *molar mass
                         = 0.1746*103.62
                         = 18.09g
Answer: 18. 09g