After 2.02 g of Al has reacted completely with 0.400 L of 2.75 M HCl (the excess reactant), what is the molarity of the remaining HCl? 2Al(s)+6HCl(aq)-->2AlCl3(aq)+3H2(g)
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To find the molarity of the remaining HCl, we first need to determine the number of moles of Al that reacted completely.
The molar mass of Al is 26.98 g/mol. Given that we have 2.02 g of Al, we can use the following equation to find the number of moles:
moles of Al = mass of Al / molar mass of Al
moles of Al = 2.02 g / 26.98 g/mol ≈ 0.075 moles of Al
According to the balanced chemical equation, Al reacts with HCl in a 2:6 ratio. This means that for every 2 moles of Al, 6 moles of HCl are required.
Since we have found that 0.075 moles of Al reacted, we can calculate the moles of HCl that reacted by using the ratio:
moles of HCl reacted = (moles of Al reacted) x (6 moles HCl / 2 moles Al)
moles of HCl reacted = 0.075 moles Al x (6 mol HCl / 2 mol Al) = 0.225 moles of HCl reacted.
Now, let's find the moles of HCl that remained after the reaction. The initial moles of HCl can be determined by using the molarity and volume information:
moles of HCl initially = molarity of HCl x volume of HCl
moles of HCl initially = 2.75 mol/L x 0.400 L = 1.10 moles of HCl initially
To find the moles of HCl remaining, we subtract the moles of HCl reacted from the initial moles of HCl:
moles of HCl remaining = moles of HCl initially - moles of HCl reacted
moles of HCl remaining = 1.10 moles of HCl initially - 0.225 moles of HCl reacted = 0.875 moles of HCl remaining
Finally, we can calculate the molarity of the remaining HCl by dividing the moles of HCl remaining by the remaining volume:
molarity of remaining HCl = moles of HCl remaining / remaining volume
molarity of remaining HCl = 0.875 moles / 0.400 L = 2.19 M
Therefore, the molarity of the remaining HCl after the reaction is 2.19 M.