Question

Aluminum reacts with excess hydrochloric acid to form aqueous aluminum chloride and 46.5 mL of hydrogen gas over water at 27 degrees Celsius and 751 mmHg. How many grams of aluminum reacted? The partial pressure of water at 27 C is 26.8 mmHg.

Answer 1

Answer: The amount of aluminium reacted is 0.0324 grams

Step-by-step explanation:

We are given:

Vapor pressure of water = 26.8 mmHg

Total vapor pressure = 751 mmHg

Vapor pressure of hydrogen gas = Total vapor pressure - Vapor pressure of water = (751- 26.8) mmHg = 724.2 mmHg

To calculate the amount of hydrogen gas collected, we use the equation given by ideal gas which follows:

`PV=nRT`

where,

P = pressure of the gas = 724.2 mmHg

V = Volume of the gas = 46.5 mL = 0.0465 L (Conversion factor: 1 L = 1000 mL)

T = Temperature of the gas =
`27^oC=[27+273]K=300K`

R = Gas constant =
`62.364\text{ L. mmHg }mol^(-1)K^(-1)`

n = number of moles of hydrogen gas = ?

Putting values in above equation, we get:

`724.2mmHg* 0.0465L=n* 62.364\text{ L.mmHg }mol^(-1)K^(-1)* 300K\\\\n=(724.2* 0.0465)/(62.364* 300)=0.0018mol`

The chemical equation for the reaction of aluminium with HCl follows:

`2Al+6HCl\rightarrow 2AlCl_3+3H_2`

By Stoichiometry of the reaction:

3 moles of hydrogen gas is produced when 2 moles of aluminium is reacted

So, 0.0018 moles of hydrogen gas will be produces when =
`(2)/(3)* 0.0018=0.0012mol` of aluminium is reacted

To calculate the mass from given number of moles, we use the equation:

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

Moles of aluminium = 0.0012 moles

Molar mass of aluminium = 27 g/mol

Putting values in above equation, we get:

`0.0012mol=\frac{\text{Mass of aluminium}}{27g/mol}\\\\\text{Mass of aluminium}=(0.0012mol* 27g/mol)=0.0324g`

Hence, the amount of aluminium reacted is 0.0324 grams