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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Note that 2 mol dm⁻³ = 2 mol/(dm³)outline: we're given the concentration of hcl, which is  2 mol dm⁻³ (cubic meters in the denominator can represent volume, so i know this is concentration). we're an amount of calcium carbonate caco₃, which is 5 g. we're presented with a chemical equation which includes caco₃ and hcl. we are told to find the volume of hcl. from what we see so far, this seems to be a stoichiometry  problem. we'll have to use the given information about calcium carbonate to find the amount of moles of hcl needed to fully react with calcium carbonate. then we'll have information about the moles of hcl and the concentration of hcl; that will be enough to figure out volume according to the formula concentration = moles / volume.======all the answer choices are not in dm.convert the given concentration so that it uses cm⁻³.we can use conversion factors. we have to cube our conversion factors so that dimensional analysis works.we can convert to meters first, using the fact that     10 dm = 1 m  ⇔ 10³ dm³ = 1 m³then we use the fact that     10² cm = 1 m ⇔ (10² cm)³ = (1 m)³ ⇔ 10⁶ cm³ = 1 m³so converting:     now we have to find the moles of hydrochloric  acid neededfind the moles of calcium carbonate. we have to find the molar mass of calcium carbonate, caco₃, which can be found by    molar mass = (mass ca) + (mass c) + 3(mass o)    = 40 + 12 + 3(16)    = 100 g/molconvert the given 5 g to moles:     mol caco₃ = 5 g caco₃ * 1 mol/(100 g) = 0.05 mol caco₃use stoichiometry (ratios) to figure out the moles of hcl needed. ratio wise, there's 2 hcls for every caco₃, so use dimensional analysis    mol hcl = mol caco₃ * (2 mol hcl) / (1 mol caco₃)    = 0.05 mol hcl * 2    = 0.1 mol hcl  we have 0.1 mol hcl and concentration of hcl (0.002 mol/cm³).we need to find volumesince concentration = moles / volume, rearrange to get     volume = moles / concentration    volume = 0.1 mol / (0.002 mol/cm³)    volume = 50 cm³