Please help

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

8 0

3 years ago

Gravity relates to a stream's ability to erode and deposit material by causing water to flow faster down a steep slope. True or

blsea [12.9K]

Answer:

true

Explanation:

4 0

2 years ago