How would you calculate the % yield of hydrogen if 40.0 grams of magnesium react with an excess of nitric acid producing 1.70 grams of hydrogen gas. mg + 2 hno3 → mg(no3)2 + h2

Answer is: yield of hydrogen is 51.5%.
Balanced chemical reaction: Mg + 2HNO₃ → Mg(NO₃)₂ + H₂.
m(Mg) = 40.0g.
n(Mg) = m(Mg) ÷ M(Mg).
n(Mg) = 40 g ÷ 24.3 g/mol.
n(Mg) = 1.65 mol.
From chemical reaction: n(Mg) : n(H₂) = 1 : 1.
n(H₂) = n(Mg) = 1.65 mol.
m(H₂) = 1.65 mol · 2 g/mol.
m(H₂) = 3.3 g.
yield = 1.70 g ÷ 3.3 g · 100% = 51.5%.
n - amount of substance.
M - molar mass.

The percentage yield of Hydrogen is 51.1%

We'll begin by calculating the mass of Mg that reacted and the mass of H₂ produced from the balanced equation.

Mg + 2HNO₃ → Mg(NO₃)₂ + H₂

Molar mass of Mg = 24 g/mol

Mass of Mg from the balanced equation = 1 × 24 = 24 g

Molar mass of H₂ = 2 × 1 = 2 g/mol

Mass of H₂ from the balanced equation = 1 × 2 = 2 g

From the balanced equation above,

24 g of Mg reacted to produce 2 g of H₂.

Next, we shall determine the theoretical yield of H₂.

From the balanced equation above,

24 g of Mg reacted to produce 2 g of H₂.

Therefore,

40 g of Mg will react to produce = (40 × 2) / 24 = 3.33 g of H₂.

Thus the theoretical yield of H₂ is 3.33 g

Finally, we shall determine the percentage yield of H₂.

Actual yield = 1.70 gTheoretical yield = 3.33 gPercentage yield =?

Percentage yield = (Actual / Theoretical) × 100

Percentage yield = (1.70 / 3.33) × 100

Percentage yield of H₂ = 51.1%

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Molarity is moles of solute over liters of solution. idk if that is what you are talking about, but you can convert moles of solute to grams of solute using dimensional analyzes by using molar mass found on the periodic table.