Question

if 0.600 mol of chlorine gas is reacted with 0.500mol of aluminium metal to produce aluminium chloride, which reactant is in excess?how many moles of aluminium chloride can be produced during the reaction ?
kindly give me the correct answer step by step ..its a humble request​

Answer 1

Answer:

I can't understand...............

Explanation:

it's mayank

Answer 2

Answer: 0.4 moles

Explanation: Balanced chemical equation

2Al(s) + 3Cl2(g)-------> 2AlCl3 (s)

From the above equation, it can be seen that 2 moles of Al combine with 3 moles of Cl and produce 2 moles of AlCl3.

2 mol of Al=2 mol of AlCl3

3 mol of Cl= 2 mol of AlCl3

Now, we are given 0.600 mol of Cl and 0.500 mol of Al.

Let's find limiting reagent

2 mol of Al= 2 mol of AlCl3

0.500 of Al= x

As we see, mol of Al and AlCl3 are same, so 0.500 mol of Al produce 0.500 mol of AlCl3.

Now for Cl

3 mol of Cl= 2 mol of AlCl3

0.600 mol of Cl=x

By cross multiplication;

2×0.600/3= 0.4

As, Al produces 0.500 mol of AlCl3 and Cl produces 0.4 mol of AlCl3, it is concluded that Al is in excess and Cl is a limiting reagent. As limiting reagent stops the reaction, so the number of moles of AlCl3 produced is;

3 mol of Cl= 2 mol of AlCl3

0.600 mol of Cl=x

2×0.600/3= 0.4 mol