Question

QUESTION 2 Given the unbalanced equation: ____ Al(s) + ____Cl2(g) → _____AlCl3(s) If 50.00 g of aluminum and 150.00 g of chlorine gas are reacted.Calculate the maximum number of moles of AlCl3 that can be formed. a)1.0579 mol b) 2.1157 mol c) 1.853 mol d)1.4104 mol e) 0.9266 mol

Answer 1

Molar mass of Al = 26.98 g/mol

mass(Al)= 50.0 g

use:

number of mol of Al,

n = mass of Al/molar mass of Al

=(50 g)/(26.98 g/mol)

= 1.853 mol

Molar mass of Cl2 = 70.9 g/mol

mass(Cl2)= 150.0 g

use:

number of mol of Cl2,

n = mass of Cl2/molar mass of Cl2

=(1.5*10^2 g)/(70.9 g/mol)

= 2.116 mol

Balanced chemical equation is:

2 Al + 3 Cl2 ---> 2 AlCl3 +

2 mol of Al reacts with 3 mol of Cl2

for 1.853 mol of Al, 2.78 mol of Cl2 is required

But we have 2.116 mol of Cl2

so, Cl2 is limiting reagent

we will use Cl2 in further calculation

According to balanced equation

mol of AlCl3 formed = (2/3)* moles of Cl2

= (2/3)*2.116

= 1.4107 mol

Answer: d