Respuesta :
Answer: The amount of aluminium reacted is 0.0324 grams
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:
[tex]PV=nRT[/tex]
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 = [tex]27^oC=[27+273]K=300K[/tex]
R = Gas constant = [tex]62.364\text{ L. mmHg }mol^{-1}K^{-1}[/tex]
n = number of moles of hydrogen gas = ?
Putting values in above equation, we get:
[tex]724.2mmHg\times 0.0465L=n\times 62.364\text{ L.mmHg }mol^{-1}K^{-1}\times 300K\\\\n=\frac{724.2\times 0.0465}{62.364\times 300}=0.0018mol[/tex]
The chemical equation for the reaction of aluminium with HCl follows:
[tex]2Al+6HCl\rightarrow 2AlCl_3+3H_2[/tex]
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 = [tex]\frac{2}{3}\times 0.0018=0.0012mol[/tex] of aluminium is reacted
To calculate the mass from given number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Moles of aluminium = 0.0012 moles
Molar mass of aluminium = 27 g/mol
Putting values in above equation, we get:
[tex]0.0012mol=\frac{\text{Mass of aluminium}}{27g/mol}\\\\\text{Mass of aluminium}=(0.0012mol\times 27g/mol)=0.0324g[/tex]
Hence, the amount of aluminium reacted is 0.0324 grams
