Question

For each of the reactions shown, calculate the mass of the product formed when 10.84 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)

2- 2K(s)+Br2(l)→2KBr(s)

3- 4Cr(s)+3O2(g)→2Cr2O3(s)

4- 2Sr(s)+O2(g)→2SrO(s)

Answer 1

1- 2K(s)+Cl₂(g)→2KCl(s)

Molar mass of Cl₂ is = 2xAt.mass of Cl = 2x35.5 = 71 g/mol

Molar mass of KCl = At.mass of K + At.mass of Cl = 39+35.5 = 74.5 g/mol

According to the balanced equation,

1 mole of Cl₂ produces 2 moles of KCl

                        OR

1x71 g of Cl₂ produces 2x74.5 g of KCl

10.84 g of Cl₂ produces M g of KCl

M = ( 10.84x2x74.5) / 71

   = 22.75 g

2- 2K(s)+Br₂(l)→2KBr(s)

Molar mass of Br₂ is = 2xAt.mass of Br = 2x80 = 160 g/mol

Molar mass of KBr = At.mass of K + At.mass of Br = 39+80 = 119 g/mol

According to the balanced equation,

1 mole of Br₂ produces 2 moles of KBr

                        OR

1x160 g of Br₂ produces 2x119 g of KBr

10.84 g of Br₂ produces M g of KBr

M = ( 10.84x2x119) / 160

   = 16.12 g

3- 4Cr(s)+3O₂(g)→2Cr₂O₃(s)

Molar mass of O₂ is = 2xAt.mass of O = 2x16 = 32 g/mol

Molar mass of Cr₂O₃= (2xAt.mass of Cr) +(3x At.mass of O) = (2x52)+(3x16) = 152 g/mol

According to the balanced equation,

3 mole of O₂ produces 2 moles of Cr₂O₃

                        OR

3x32 g of O₂ produces 2x152 g of Cr₂O₃

10.84 g of O₂ produces M g of Cr₂O₃

M = ( 10.84x2x152) / (3x32)

   = 34.33 g

4- 2Sr(s)+O₂(g)→2SrO(s)

Molar mass of Sr s = 87.6 g/mol

Molar mass of SrO = At.mass of Sr + At.mass of O = 87.6+16 = 103.6 g/mol

According to the balanced equation,

2 mole of Sr produces 2 moles of SrO

                        OR

2 x 87.6 g of Sr produces 2 x103.6 g of SrO

10.84 g of Sr produces M g of SrO

M = ( 10.84x2x103.6) / (2x87.6)

   = 12.82 g