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Rudik [331]
3 years ago
9

s) Suppose we now collect hydrogen gas, H2(g), over water at 21◦C in a vessel with total pressure of 743 Torr. If the hydrogen g

as is produced by the reaction of aluminum with hydrochloric acid: 2Al(s) + 6HCl(aq) → 2AlCl3(aq) + 3H2(g) what volume of hydr
Chemistry
1 answer:
Elenna [48]3 years ago
8 0

This is an incomplete question, here is a complete question.

Suppose we now collect hydrogen gas, H₂(g), over water at 21°C in a vessel with total pressure of 743 Torr. If the hydrogen gas is produced by the reaction of aluminum with hydrochloric acid:

2Al(s)+6HCl(aq)\rightarrow 2AlCl_3(aq)+3H_2(g)

what volume of hydrogen gas will be collected if 1.35 g Al(s) reacts with excess HCl(aq)? Express  your answer in liters.

Answer : The volume of hydrogen gas that will be collected is 1.85 L

Explanation :

First we have to calculate the number of moles of aluminium.

Given mass of aluminium = 1.35 g

Molar mass of aluminium = 27 g/mol

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

\text{Moles of aluminium}=\frac{1.35g}{27g/mol}=0.05mol

The given chemical reaction is:

2Al(s)+6HCl(aq)\rightarrow 2AlCl_3(aq)+3H_2(g)

As, hydrochloric acid is present in excess. So, it is considered as an excess reagent.

Thus, aluminium is a limiting reagent because it limits the formation of products.

By Stoichiometry of the reaction:

2 moles of aluminium produces 3 moles of hydrogen gas

So, 0.005 moles of aluminium will produce = \frac{3}{2}\times 0.05=0.0750mol of hydrogen gas

Now we have to calculate the mass of helium gas by using ideal gas equation.

PV = nRT

where,

P = Pressure of hydrogen gas = 743 Torr

V = Volume of the helium gas = ?

n = number of moles of hydrogen gas = 0.075 mol

R = Gas constant = 62.364\text{ L Torr }mol^{-1}K^{-1}

T = Temperature of hydrogen gas = 21^oC=[21+273]K=294K

Now put all the given values in above equation, we get:

743Torr\times V=0.075mol\times 62.364\text{ L Torr }mol^{-1}K^{-1}\times 294K\\\\V=1.85L

Hence, the volume of hydrogen gas that will be collected is 1.85 L

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