Question

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 number of moles of Al required to react with all the Cl2 present. a)3.1736 mol b) 1.4104 mol c)1.853 mol d)1.0579 mol e)2.1157 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

So,

mol of Al reacted = (2/3)*moles of Cl2

= (2/3)*2.116 mol

= 1.4107 mol

Answer: b