Al(s) + HCl(aq) → AlCl3(aq) + H2(g) According to the equation above, how many grams of aluminum are needed to completely react with 2.19 mol of hydrochloric acid?
I can give an answer "according to the equation above" but it won't mean anything because the equation above is not balanced. Let me balance it and answer correctly.
2Al + 6 HCl ==> 2AlCl3 + 3H2
2.19 moles HCl will require 2.19 x (2 moles Al/6 moles HCl) = 2.19*2/6 = 0.73 moles Al. Convert that to grams Al; i.e.., grams Al = mols Al x atomic mass Al = ?
To find out the number of grams of aluminum needed to completely react with 2.19 mol of hydrochloric acid, we need to use the mole ratio between aluminum and hydrochloric acid from the balanced equation.
From the balanced equation: 1 mol of aluminum reacts with 3 mol of hydrochloric acid.
First, we need to convert the given amount of hydrochloric acid to moles:
2.19 mol HCl x (1 mol AlCl3 / 3 mol HCl) = 0.73 mol AlCl3
Next, we can use the molar mass of aluminum to convert moles of aluminum chloride to grams of aluminum:
Molar mass of aluminum (Al) = 26.98 g/mol
0.73 mol Al x (26.98 g Al / 1 mol Al) = 19.66 g Al
Therefore, approximately 19.66 grams of aluminum are needed to completely react with 2.19 mol of hydrochloric acid.
To determine the grams of aluminum needed to completely react with 2.19 mol of hydrochloric acid, we need to first balance the equation:
2Al(s) + 6HCl(aq) → 2AlCl3(aq) + 3H2(g)
According to the balanced equation, it shows that 2 moles of aluminum are required to react with 6 moles of hydrochloric acid to produce 2 moles of aluminum chloride and 3 moles of hydrogen gas.
From the balanced equation, the mole ratio of aluminum to hydrochloric acid is 2:6 or 1:3. This means that for every mole of aluminum, 3 moles of hydrochloric acid are consumed.
Given that there are 2.19 mol of hydrochloric acid, we can use the mole ratio to calculate the moles of aluminum needed.
2.19 mol HCl * (1 mol Al / 3 mol HCl) = 0.73 mol Al
Now, to convert the moles of aluminum to grams, we need to use the molar mass of aluminum. The molar mass of aluminum (Al) is 26.98 g/mol.
0.73 mol Al * 26.98 g/mol = 19.65 grams of aluminum.
Therefore, approximately 19.65 grams of aluminum are needed to completely react with 2.19 mol of hydrochloric acid.