5.4 g of Aluminium reacts with 300 mL of 0.2 mol/L hydrochloric acid solution. A. Write equation for the reaction taking place. B. Specify which reactant is limiting and which reactant is excess. C. Find volume of the gas collected at S. T.P d. How many grams of salt are produced at the end of the reaction? E. How many grams of the excess reactant are left ate the end of the reaction? Given: Al=27 , H=1 , Cl=35.5 Chemistry grade 10

A. 2 Al (s) + 6HCl (aq) →  2AlCl₃ (s) ↓ + 3H₂ (g)  

B. Al is the excess reactant and HCl is the limiting.

C. 0.672 L of H₂ produced at STP

D. 2.67 g of AlCl₃ are made in this reaction.

E. 4.86 g of Al remain after the reaction goes complete.

Explanation:

We star from the reaction:

2 Al (s) + 6HCl (aq) →  2AlCl₃ (s) ↓ + 3H₂ (g)  

2 moles of aluminum, react with 6 moles of HCl in order to produce 2 moles of aluminum chloride and 3 mol of H₂ gas.

We determine moles of each reactant:

[HCl] = 0.2M → 0.2 mol/L . 0.3L = 0.060 moles

(we converted 300 mL to 0.3L)

5.4 g of Al . 1mol / 26.98g = 0.200 moles

Ratio is 2:6 (3). 2 mol of Al react to 6 mol of HCl

0.2 moles of Al may react with (0.2 . 6) /2 = 0.6 mol of acid

We have 0.06 moles, and we need 0.6 mol of acid, so the HCl is the limiting reactant. Then, the Al is the excess:

6 moles of HCl need 2 moles of Al to react

Then 0.06 moles of HCl will react to (0.06 . 2) /6 = 0.02 moles

If we have 0.2 moles of Al, and we need 0.02 moles for the reaction, then

(0.2 - 0.02) = 0.18 moles remain after the reaction is complete.

0.18 mol . 26.98g /1mol = 4.86 g of Al remain after the reaction goes complete.

As the limting reactant is the HCl, we work with it to determine the mass of salt which is produced:

6 mol of HCl can produce 2 mol of chloride

Then 0.06 moles of HCl will produce (0.06 . 2) /6 = 0.02 mol of AlCl₃

We convert to mass: 0.02 mol . 133.33g/1mol = 2.67 g of AlCl₃ are made in this reaction.

Let's find out the volume of hydrogen produced, at STP

6 moles of HCl can produce 3 moles of H₂

0.06 moles of HCl will produce (0.06 . 3) /6 = 0.03 moles of H₂

1 mol of any gas at STP occupies 22.4L

0.03 moles of H₂ will ocuppy (22.4 L . 0.03 mol)/1mol = 0.672L

the answer is   c.) 1575 j.

explanation: