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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:
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
The given chemical reaction is:
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 = 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 =
T = Temperature of hydrogen gas =
Now put all the given values in above equation, we get:
Hence, the volume of hydrogen gas that will be collected is 1.85 L
%Mass
Ar C = 12 g/mol, Mr C₄H₁₀ = 58 g/mol, Ar H = 1 g/mol
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