Question

How many grams of NiCl2 can be formed in the reaction of 3.76 g of Ni(s) and 1.98 g of HCl? The other product is H2(g). Report your answer to the hundredths of a gram without units.

Answer 1

Molar mass of Ni = 58.69 g/mol

mass(Ni)= 3.76 g

use:

number of mol of Ni,

n = mass of Ni/molar mass of Ni

=(3.76 g)/(58.69 g/mol)

= 6.407*10^-2 mol

Molar mass of HCl,

MM = 1*MM(H) + 1*MM(Cl)

= 1*1.008 + 1*35.45

= 36.458 g/mol

mass(HCl)= 1.98 g

use:

number of mol of HCl,

n = mass of HCl/molar mass of HCl

=(1.98 g)/(36.46 g/mol)

= 5.431*10^-2 mol

Balanced chemical equation is:

Ni + 2 HCl ---> NiCl2 + H2

1 mol of Ni reacts with 2 mol of HCl

for 6.407*10^-2 mol of Ni, 0.1281 mol of HCl is required

But we have 5.431*10^-2 mol of HCl

so, HCl is limiting reagent

we will use HCl in further calculation

Molar mass of NiCl2,

MM = 1*MM(Ni) + 2*MM(Cl)

= 1*58.69 + 2*35.45

= 129.59 g/mol

According to balanced equation

mol of NiCl2 formed = (1/2)* moles of HCl

= (1/2)*5.431*10^-2

= 2.715*10^-2 mol

use:

mass of NiCl2 = number of mol * molar mass

= 2.715*10^-2*1.296*10^2

= 3.519 g

Answer: 3.52 g