Question

The following reaction describes the formation of nickel tetracarbonyl:

Ni(s) + CO(g) --> Ni(CO)4(g)

Consider a reaction which produces 12.4 g of nickel tetracarbonyl.

a) balance the reaction equation

b) Find the number of grams of Ni(s) and CO(g) which reacted.

Answer 1

a)

Ni(s) + 4 CO (g) —> Ni(CO)4 (g)

b)

Molar mass of Ni(CO)4 = 1*MM(Ni) + 4*MM(C) + 4*MM(O)

= 1*58.69 + 4*12.01 + 4*16.0

= 170.73 g/mol

mass of Ni(CO)4 = 12.4 g

mol of Ni(CO)4 = (mass)/(molar mass)

= 12.4/170.73

= 0.0726 mol

From balanced chemical reaction, we see that

mol of Ni reacted = moles of Ni(CO)4

= 0.0726 mol

Molar mass of Ni = 58.69 g/mol

mass of Ni = number of mol * molar mass

= 0.0726*58.69

= 4.26 g

Answer: mass of Ni(s) = 4.26 g

mol of CO reacted = (4/1)* moles of Ni(CO)4

= (4/1)*0.0726

= 0.2905 mol

Molar mass of CO = 1*MM(C) + 1*MM(O)

= 1*12.01 + 1*16.0

= 28.01 g/mol

mass of CO = number of mol * molar mass

= 0.2905*28.01

= 8.14 g

Answer: mass of CO(g) = 8.14 g