Question

4. How many grams of hydrogen is produced from 12.5 g of Mg reacting with hydrochloric acid in this balanced equation?

Mg + 2HCl ---> MgCl2 + H2

Answer 1

To find out the mass of hydrogen produced when 12.5 g of magnesium (Mg) reacts with hydrochloric acid (HCl), we can use stoichiometry. Here is how to approach the problem: Step 1: Write the balanced chemical equation: Mg + 2HCl → MgCl2 + H2The balanced equation shows that 1 mole of magnesium reacts with 2 moles of hydrochloric acid to produce 1 mole of hydrogen gas. Step 2: Calculate the number of moles of Mg in 12.5 g of Mg using the molar mass of Mg12.5 g Mg × (1 mol Mg/24.31 g Mg) = 0.514 mol Mg Step 3: Use stoichiometry to find the number of moles of H2 produced by reacting 0.514 mol of Mg with HCl0.514 mol Mg × (1 mol H2/1 mol Mg) = 0.514 mol H2Step 4: Calculate the mass of hydrogen gas produced by multiplying the number of moles of H2 with its molar mass0.514 mol H2 × (2.016 g H2/1 mol H2) = 1.035 g H2 Therefore, the mass of hydrogen produced when 12.5 g of magnesium reacts with hydrochloric acid in the balanced equation is 1.035 g H2.

Answer 2

The balanced equation shows that 1 mole of Mg reacts with 2 moles of HCl to produce 1 mole of H2.

First, we need to find the number of moles of Mg in 12.5 g of Mg:

12.5 g Mg / 24.31 g/mol = 0.514 moles Mg

Next, we can use the mole ratio from the balanced equation to find the number of moles of H2 produced:

0.514 moles Mg x (1 mole H2 / 1 mole Mg) = 0.514 moles H2

Finally, we can convert the number of moles of H2 to grams:

0.514 moles H2 x 2.016 g/mol = 1.036 g H2

Therefore, 1.036 grams of hydrogen is produced from 12.5 g of Mg reacting with hydrochloric acid in this balanced equation.