in 25grams of aluminium oxide reacts with 50grams aquoues hyrdogen chloride what is the mass of aluminum chloride produced?
This is a limiting reagent problem. You know a limiting reagent problem when you see it because it always lists BOTH of the reactants.
1. Write and balance the equation.
2. Convert 25 g Al2O3 and 50 g HCl to moles. moles = grams/molar mass.
3a. Using the coefficients in the balanced equation, convert moles Al2O3 to moles AlCl3.
3b. Using the same procedure, convert moles HCl to moles AlCl3.
3c. You have two answers for the same product. Obviously, one of them is incorrect. The correct value is ALWAYS the smaller one and that is the limiting reagent.
4. Now convert the correct value from 3c to grams. grams = moles x molar mass.
Post your work if you get stuck.
i got the mass of aluminum chloride to be 42.7
I didn't obtain that number. Post your work and I'll try to find the error. Start with the balanced equation
Al2O3 + 6HCl ==> 2AlCl3 + 3H2O
To calculate the mass of aluminum chloride produced from the reaction between aluminum oxide and hydrogen chloride, first, we need to write the balanced chemical equation for the reaction:
2Al2O3 + 6HCl → 2AlCl3 + 3H2O
From the equation, we see that 2 moles of aluminum oxide (Al2O3) react with 6 moles of hydrochloric acid (HCl) to produce 2 moles of aluminum chloride (AlCl3).
To find the mass of aluminum chloride produced, we need to determine the limiting reactant, which is the reactant that is completely consumed in the reaction. This can be done by comparing the number of moles of each reactant to their stoichiometric coefficients in the balanced equation.
Let's calculate the number of moles of aluminum oxide (Al2O3):
Molar mass of Al2O3:
2(27 g/mol for Al) + 3(16 g/mol for O) = 102 g/mol
Number of moles of Al2O3:
25 g / 102 g/mol = 0.245 moles
Now, let's calculate the number of moles of hydrochloric acid (HCl):
Molar mass of HCl: 1(1 g/mol for H) + 1(35.5 g/mol for Cl) = 36.5 g/mol
Number of moles of HCl:
50 g / 36.5 g/mol ≈ 1.37 moles
Since the balanced equation shows that 2 moles of Al2O3 react to produce 2 moles of AlCl3, we can infer that 0.245 moles of Al2O3 will produce 0.245 moles of AlCl3.
Therefore, the mass of aluminum chloride produced can be calculated by multiplying the number of moles of AlCl3 by its molar mass:
Molar mass of AlCl3: 1(27 g/mol for Al) + 3(35.5 g/mol for Cl) = 133.5 g/mol
Mass of AlCl3:
0.245 moles * 133.5 g/mol ≈ 32.7 grams
So, approximately 32.7 grams of aluminum chloride will be produced from the given reaction.