Assuming a yield of 26.10 percent calculate the actual yield of magnesium nitrate formed from 42.2 g of magnesium and excess copper(ii) nitrate .
molar mass Mg ... 24.3 g
molar mass magnesium nitrate ... 148.3 g
.261 * (42.2 / 24.3) * 148 g
To calculate the actual yield of magnesium nitrate, we need to determine the amount of magnesium nitrate that should theoretically be formed and then multiply it by the percent yield.
1. Start by writing a balanced equation for the reaction between magnesium and copper(II) nitrate:
3Mg + 2Cu(NO3)2 --> 3Mg(NO3)2 + 2Cu
2. Determine the molar masses of magnesium and magnesium nitrate:
Molar mass of Mg = 24.31 g/mol
Molar mass of Mg(NO3)2 = 148.33 g/mol
3. Convert the mass of magnesium (42.2 g) to moles:
Moles of Mg = Mass of Mg / Molar mass of Mg
Moles of Mg = 42.2 g / 24.31 g/mol
4. Use the stoichiometry of the balanced equation to convert moles of Mg to moles of Mg(NO3)2:
From the balanced equation, we see that 3 moles of Mg produce 3 moles of Mg(NO3)2.
Moles of Mg(NO3)2 = Moles of Mg
5. Calculate the theoretical yield of magnesium nitrate in grams:
Theoretical yield = Moles of Mg(NO3)2 × Molar mass of Mg(NO3)2
6. Now, we can calculate the actual yield using the given percent yield:
Actual yield = Theoretical yield × (Percent yield / 100)
Actual yield = Theoretical yield × (26.10 / 100)
Where the Percent yield is given as 26.10%.
By following these steps, you can find the actual yield of magnesium nitrate formed from 42.2 g of magnesium and excess copper(II) nitrate.