Answer:
The reaction requires 0.795 grams of HCl. True.
This reaction also produces 1.04 grams of MgCl₂. True.
The number of moles of the reactants consumed will equal the number of moles of the products made. False.
Explanation:
- For the balanced chemical reaction:
<em>2HCl(aq) + Mg(s) → MgCl₂(aq) + H₂(g),</em>
<em>It is clear that 2 mol of HCl react with 1 mol of Mg to produce 1 mol of MgCl₂ and 2 mol of H₂.</em>
<em />
<em>The reaction requires 0.795 grams of HCl.</em>
- Firstly, we need to calculate the no. of moles of produced H₂ (0.022 g) using the relation:
no. of moles of H₂ = mass/molar mass = (0.022 g)/(2.015 g/mol) = 0.01092 mol.
- To find the required mass of HCl to produce 0.022 g of H₂ (0.01092 mol):
<em><u>using cross multiplication:</u></em>
2 mol of HCl produce → 1 mol of H₂, from stichiometry.
??? mol of HCl produce → 0.01092 mol of H₂.
∴ The no. of moles of HCl needed to produce (0.01092 mol) of H₂ = (2 mol)(0.01092 mol)/(1 mol) = 0.02184 mol.
∴ The mass of HCl needed = no. of moles * molar mass = (0.02184 mol)*(36.46 g/mol) = 0.796 g.
<em>So, this statement is true.</em>
<em></em>
<em>This reaction also produces 1.04 grams of MgCl₂.</em>
- To find the mass of MgCl₂ produced with 0.022 g of H₂ (0.01092 mol):
<u><em>using cross multiplication:</em></u>
1 mol of MgCl₂ produced with → 1 mol of H₂, from stichiometry.
0.01092 mol of MgCl₂ produce with → 0.01092 mol of H₂.
∴ The mass of MgCl₂ produced = no. of moles * molar mass = (0.01092 mol)*(95.211 g/mol) = 1.04 g.
<em>So, this statement is true.</em>
<em></em>
<em>The number of moles of the reactants consumed will equal the number of moles of the products made.</em>
From the stichiometry 3 moles of reactants (2 mol of HCl, 1 mol of Mg) are reacted to produce 2 moles of products (1 mol of MgCl₂, 1 mol of H₂).
<em></em>
<em>So, the statement is false.</em>
