Aluminum reacts with chlorine gas to form aluminum chloride via the following re

Question

Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction:

2Al(s)+3Cl2(g)?2AlCl3(s)

You are given 12.0 g of aluminum and 17.0 g of chlorine gas.

If you had excess aluminum, how many moles of aluminum chloride could be produced from 17.0 g of chlorine gas, Cl2?

In the following chemical reaction, 2 mol of A will react with 1 mol of B to produce 1 mol of A2B without anything left over But what if you're given 2.8 mol of A and 3.2 mol the reactant that runs out first, called the limiting reactant. To identify the limiting reactant, calculate the amount of product formed from each amount of reactant separately 1.4 mol A2 amount of reactant A. Thus, A is the limiting reactant, and a maximum of1 be formed from the given amounts.

Explanation / Answer

Molar mass of Al = 27 gm

mass of Al= 12g

number of moles of Al = mass/molar mass = 12/27 = 0.44 mol

Molar mass of Cl2 = 71 gm

mass of Cl2 = 17 gm

number of moles of Cl2 = mass/molar mass = 17/71 = 0.24 mol

Now 2mol of Al requires 3mol of Cl2, so Cl2 is present in less quantity and hence it is limiting reagent.

3 mol of Cl2 produces 2 mol ofAlCl3,

number of moles of AlCl3 produced= (2/3)* number of moles of Cl2

                 = (2/3)*0.24

                  = 0.16mol

Answer:0.16 mol

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