Question

For each of the reactions, calculate the mass (in grams) of the product formed when 15.45 g of the underlined reactant completely reacts. Assume that there is more than enough of the other reactant.

2K(s)+Cl2(g)−−−−−→2KCl(s)​

2K(s)+Br2(l)−−−−−→2KBr(s)​

4Cr(s)+3O2(g)−−−−−→2Cr2O3(s)

2Sr(s)−−−−+O2(g)→2SrO(s)​

Answer 1

(a) Given

Mass of CL₂ = 15.45 g

Molar mass of Cl₂ = 70.90 g/mol

Moles of Cl₂ = mass/ molar mass = 15.45 g/(70.90g/mol) = 0.2179 mol

1 mol of Cl₂ given 2 mol of KCl

Thus,02179 mol of  Cl₂ given 0.4358 mol of KCl

Molar mass of KCl = 74.55 g/mol

Thus, Mass of KCl formes = moles * Molar mass = 0.4358 mol*74.55 g/mol = 32.49 g

(B)

Mass of Br₂ = 15.45 g

Molar mass of Br₂ = 159.8 g/mol

Moles of Br₂ = mass/ molar mass = 15.45 g/(159.8g/mol) = 0.0967 mol

1 mol of Br₂ given 2 mol of KBr

Thus,0.0967 mol of Br₂ given 0.1934 mol of KBr

Molar mass of KBr = 119.0 g/mol

Thus, Mass of KBr formes = moles * Molar mass = 0.1934 mol*119.0 g/mol = 23.01 g

(C)

Mass of O₂ = 15.45 g

Molar mass of O₂ = 32 g/mol

Moles of O₂ = mass/ molar mass = 15.45 g/(32g/mol) = 0.4828 mol

3 mol of O₂ given 2 mol of C₂O₃

Thus,0.4828 mol of O₂ given 0.3219 mol of C₂O₃​

Molar mass of C₂O₃​ = 152 g/mol

Thus, Mass of C₂O₃​ formes = moles * Molar mass = 0.3219 mol*152g/mol = 48.93 g

(D)Mass of Sr = 15.45 g

Molar mass of Sr = 87.62 g/mol

Moles of Sr = 15.45 g/(87.62g/mol) = 0.1763 mol

2 mol of Sr give 2 mol of SrO

THus, Moles of SrO formed = 0.1763 mol

Molar mass of SrO = 103.62 g/mol

Mass of SrO formed = moles* molar mass = 0.1763 mol* 103.62 g/mol = 18.27 g