Question

Consider the following reaction carried out under constant pressure 6HCl(aq) + 2Al(s) → 3H2(g) + 2AlCl3(s) Δ Hrxn = -4.04×102 kJ Calculate the heat associated with the complete reaction of 4.38×102 g of HCl with 67.0 g of Al. Show all work.

A.-4.85×103 kJ

B.-3.96×102 kJ

C.-8.08×102 kJ

D.-5.01×102 kJ

E.-1.00×103 kJ

Answer 1

Molar mass of HCl = 1*MM(H) + 1*MM(Cl)

= 1*1.008 + 1*35.45

= 36.458 g/mol

mass of HCl = 438.0 g

we have below equation to be used:

number of mol of HCl,

n = mass of HCl/molar mass of HCl

=(438.0 g)/(36.458 g/mol)

= 12.01 mol

Molar mass of Al = 26.98 g/mol

mass of Al = 67.0 g

we have below equation to be used:

number of mol of Al,

n = mass of Al/molar mass of Al

=(67.0 g)/(26.98 g/mol)

= 2.483 mol

we have the Balanced chemical equation as:

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

6 mol of HCl reacts with 2 mol of Al

for 12.01 mol of HCl, 4.005 mol of Al is required

But we have 2.483 mol of Al

so, Al is limiting reagent

we will use Al in further calculation

Since delta H is negative, heat is released

when 2 mol of Al reacts, heat released = 404.0 KJ

So,

for 2.483 mol of Al, heat released = 2.483*404.0/2 KJ

= 5.01*10^2 KJ

Answer: - 5.01*10^2 KJ

Answer: D