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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

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Dejanras
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.
Eduard22sly

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 g
  • Theoretical yield = 3.33 g
  • Percentage yield =?

Percentage yield = (Actual / Theoretical) × 100

Percentage yield = (1.70 / 3.33) × 100

Percentage yield of H₂ = 51.1%

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