Consider 0.022 grams of h2(g) produced by the following chemical reaction, 2 hcl(aq) + mg(s) → mgcl2(aq) + h2(g) determine if each of the following statements is true or false. click here for a copy of the periodic table. the reaction requires 0.795 grams of hcl. this reaction also produces 1.04 grams of mgcl2. the number of moles of the reactants consumed will equal the number of moles of the products made.

Out of the given statements, True are:

The reaction requires 0.795 grams of HCl.

This reaction also produces 1.04 grams of MgCl₂.

Out of the given statements, False are:

The number of moles of the reactants consumed will equal the number of moles of the products made.

2HCl(aq) + Mg(s) → MgCl₂(aq) + H₂(g),

When 2 mol of HCl react with 1 mol of Mg to produce 1 mol of MgCl₂ and 2 mol of H₂.

The reaction requires 0.795 grams of HCl.

Calculation for number of moles:

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):

2 mol of HCl produce → 1 mol of H₂, from stoichiometry.

??? mol of HCl produce → 0.01092 mol of H₂.

Thus, the no. of moles of HCl needed to produce (0.01092 mol) of H₂ = (2 mol)(0.01092 mol)/(1 mol) = 0.02184 mol.

And , the mass of HCl needed = no. of moles * molar mass = (0.02184 mol)*(36.46 g/mol) = 0.796 g.

So, this statement is true.

This reaction also produces 1.04 grams of MgCl₂.

To find:

The mass of MgCl₂ produced with 0.022 g of H₂ (0.01092 mol):

1 mol of MgCl₂ produced with → 1 mol of H₂, from stoichiometry.

0.01092 mol of MgCl₂ produce with → 0.01092 mol of H₂.

Thus, the mass of MgCl₂ produced = no. of moles * molar mass = (0.01092 mol)*(95.211 g/mol) = 1.04 g.

So, this statement is true.

Thus, the true statements are given above.

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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:

2HCl(aq) + Mg(s) → MgCl₂(aq) + H₂(g),

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.

The reaction requires 0.795 grams of HCl.

no. of moles of H₂ = mass/molar mass = (0.022 g)/(2.015 g/mol) = 0.01092 mol.

using cross multiplication:

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.

So, this statement is true.

This reaction also produces 1.04 grams of MgCl₂.

using cross multiplication:

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.

So, this statement is true.

The number of moles of the reactants consumed will equal the number of moles of the products made.

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).

So, the statement is false.