This is a limiting reagent (LR) problem; you know that because amounts are given for BOTH reactants. The following steps will work almost any LR problem.
1. Write and balance the equation.
Mg + 2HCl ==> MgCl2 + H2
2. Convert reactants to mols
a. mols HCl = M x L = 0.5 x 0.400 = approx 0.2
b. mols Mg = grams/atomic mass = 5/24.3 = approx 0.21
3a. Using the coefficients in the balanced equation, convert mols HCl to mols of EITHER product. I'll choose H2.
0.2 mols Mg x (1 mol H2/1 mol Mg) = 0.2 x 1/1 = 0.2 mols H2 (if we used all of it and had all of the HCl needed to react completely).
3b. Do the same and convert mols HCl to mols H2 (use H2 since I used that in 3a).
0.21 mols HCl x (1 mol H2/2 mols HCl) = 0.21 x 1/2 = 0.105 mols H2 produced if we used all of the HCl and had all of the Mg we needed.
3c. You can see that the mols H2 produced is not the same; the correct answer in LR problems is ALWAYS the smaller value and the reagent responsible for the value is called the LR.
3d. So HCl is the LR and Mg is the excess reagent.
4. You can at this point calculate how much H2 is produced. g H2 = mols H2 x molar mass H2. The problem did not ask for this.
5. The problem asked for how much Mg was used. You know all of the HCl was used so
0.105 mols HCl x (1 mol Mg/2 mols HCl) = 0.105 x 1/2 = approx 0.05 mols Mg used.
Be sure to go back through this and confirm each calculation I've done. I've estimated and rounded off here and there.
mols Mg =
400 centimetre cube of 0.5mol/decimetre cube hydrochloric acid was reacted with 5 grams of magnesium oxide.
(A) Calculate & identify the Excess & limiting reactants.
(B) What mass of magnesium actually took part in the reaction?
1 answer
2 answers