Question

4. Calculation of theoretical yield and percent yield You balanced this reaction earlier: Mg(s) + HCl (aq) H2 (g) + MgCl2 (aq) You used an excess amount of HCl (aq) for the reaction, therefore Mg is your limiting reactant. From the limiting reactant, calculate the theoretical yield of H2(g) you expected to get.

Answer 1

Mg(s) + 2 HCl (aq) → H₂(g) + MgCl₂

0.415g of H₂(g) -Assuming mass of Mg(s) = 10.0g-

Step-by-step explanation:

Balancing the reaction:

Mg(s) + HCl (aq) → H₂(g) + MgCl₂

There are in products two atoms of H and Cl, the balancing equation is:

Mg(s) + 2 HCl (aq) → H₂(g) + MgCl₂

Assuming you add 10g of Mg(s) -Limiting reactant-

10g of Mg are (Atomic mass: 24.305g/mol):

10g × (1 mol / 24.305g) = 0.411 moles of Mg

-Theoretical yield is the amount of product you would have after a chemical reaction occurs completely-

Assuming theoretical yield, as 1 mole of Mg(s) produce 1 mole of H₂(g), theoretical yield of H₂(g) is 0.411moles H₂(g). In grams:

0.411mol H₂(g) × (1.01g / mol) = 0.415g of H₂(g)