Question

For each reaction, calculate the mass of the product that forms when 14.9 g of the reactant in BOLD completely reacts. Assume that there is more than enough of the other reactant

a)2K(s)+Cl2(g)→2KCl(s)

b)2K(s)+Br2(l)→2KBr(s)

c)4Cr(s)+3O2(g)→2Cr2O3(s)

d)2Sr(s)+O2(g)→2SrO(s)

Answer 1

a) 2K (s) + Cl₂ (g)  -----> 2KCl (s)

Molar mass of Cl₂ = 71 g /mol

14.9 grams of Cl₂ = 14.9 /71 moles = 0.21 moles

Now, 1 mole of Cl₂ gives 2 moles of KCL

So, 0.21 mol will give 0.42 mol of KCl

Molar mass of KCl = (35.5 + 39) grams = 74.5 grams

So, Mass of KCl produced = 0.42*74.5 grams ( Molar mass * No. of moles ) = 31.29 grams

b) 1 mole of Br₂ produces 2 moles of KBr

Molar mass of Br₂ = 160 g /mol

Now, 14.9 grams of Br₂ = 14.9 / 160 mol = 0.093125 mol

which will produce 2*0.093125 moles of KBr = 0.18625 mole

Mass of KBr produced = 0.18625*Molar mass of KBr = 0.18625*119 = 22.16375 grams

c) 3 moles of O₂ produces 2 moles of Cr₂O₃

Hence 1 mol produces 2/3 moles of Cr₂O₃

14,9 grams of O₂ = 14.9 / 32 mol = 0.465625 (molar mass of O₂ = 32)

Hence , 0.465625 mole of O₂ produces 0.465625*(2/3) = 0.31 mol of Cr₂O₃

Now, Molar mass of Cr₂O₃ = 152 g/mol

Hence mass of product poroduced = 47.12 grams

d) 2 moles of Sr produce two moles of SrO

Hence 1 mole of Sr produce 1 mole of SrO

Molar mass of Sr = 87.62 grams /mol

Hence 14.9 grams = 14.9 / 87.62 moles = 0.17 mol

So, 1 mol of SrO is produced

Molar mass of SrO = 103.62 grams

So, 0.17 mol will produce 0.17*103.62 grams = 17.6154 grams