# Assuming an efficiency of 38.60%, calculate the actual yield of magnesium nitrate formed from 145.9 g of magnesium and excess copper(II) nitrate.

I got 230.5g but it isn't right..
I got 6 mols of Mg then calculated molar mass of Mg(NO3)2 and got 148.31 and then multiplied the two together
Then I used the percent yield given to find the actual yield.

## Mg + Cu(NO3)2 ==> Mg(NO3)2 + Cu

mols Mg = 145.9/24.3 = 6.003
The equation is 1 mol to 1 mol; therefore, you will have 6.003 mol Mg(NO3)2 formed. Molar mass I have is 148.31 to give g Mg(NO3)2 =890.29g at 100% yield.
890.29 x 0.3860 = 343.65g actual yield. You're allowed 4 s.f. and I would round to 343.6g.

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## To calculate the actual yield of magnesium nitrate formed, you need to consider the stoichiometry of the reaction and the given efficiency. Let's follow the steps:

1. Determine the balanced chemical equation for the reaction between magnesium and copper(II) nitrate:
Mg + Cu(NO3)2 -> Mg(NO3)2 + Cu

2. Calculate the molar mass of magnesium nitrate (Mg(NO3)2):
Mg: 24.31 g/mol
N: 14.01 g/mol
O: 16.00 g/mol

Molar mass of Mg(NO3)2 = (1 × Mg) + (2 × N) + (6 × O) = 24.31 + (2 × 14.01) + (6 × 16.00) = 148.31 g/mol

3. Calculate the theoretical yield of magnesium nitrate using the stoichiometry of the balanced equation:
Since there is an excess of copper(II) nitrate, the limiting reactant is magnesium. Convert the mass of magnesium (145.9 g) to moles by dividing it by the molar mass of magnesium (24.31 g/mol).

Moles of Mg = 145.9 g / 24.31 g/mol = 6.00 mol

According to the balanced equation, 1 mole of magnesium reacts with 1 mole of magnesium nitrate. Therefore, the theoretical yield of magnesium nitrate is also 6.00 mol.

4. Calculate the actual yield using the given efficiency (38.60%):
Actual Yield = Theoretical Yield × Efficiency
Actual Yield = 6.00 mol × 0.3860 = 2.316 mol

5. Finally, convert the actual yield from moles to grams by multiplying it by the molar mass of magnesium nitrate:
Actual Yield in grams = 2.316 mol × 148.31 g/mol = 343.17 g

Therefore, the actual yield of magnesium nitrate formed from 145.9 g of magnesium and excess copper(II) nitrate is 343.17 g.

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## To calculate the actual yield of magnesium nitrate formed, you will need to use the concept of stoichiometry and the given efficiency.

First, let's calculate the number of moles of magnesium (Mg) present in 145.9 g using its molar mass (24.31 g/mol):
Number of moles of Mg = Mass of Mg / Molar mass of Mg
= 145.9 g / 24.31 g/mol
≈ 6.004 mol (rounded to 4 decimal places)

Since there is excess copper(II) nitrate, all the magnesium will react with copper(II) nitrate to form magnesium nitrate.

Next, let's determine the stoichiometry of the balanced chemical equation for the reaction between magnesium and copper(II) nitrate:
Mg + Cu(NO3)2 → Mg(NO3)2 + Cu

From the balanced equation, we can see that one mole of magnesium reacts with one mole of copper(II) nitrate to produce one mole of magnesium nitrate. Therefore, the number of moles of magnesium nitrate formed will also be 6.004 mol.

Now, let's calculate the molar mass of magnesium nitrate (Mg(NO3)2):
Molar mass of Mg(NO3)2 = (24.31 g/mol × 1) + (14.01 g/mol × 2 × 3) + (16.00 g/mol × 8)
= 148.31 g/mol

Finally, to calculate the actual yield, multiply the number of moles of magnesium nitrate formed by its molar mass:
Actual yield = Number of moles of Mg(NO3)2 × Molar mass of Mg(NO3)2
= 6.004 mol × 148.31 g/mol
≈ 890.029g (rounded to 3 decimal places)

Therefore, the actual yield of magnesium nitrate formed is approximately 890.029 grams, rather than the value you obtained.