Assuming an efficiency of 30.80%, calculate the actual yield of magnesium nitrate formed from 147.4 g of magnesium and excess copper(II) nitrate.
Mg+Cu(NO3)2=Mg(NO3)2+Cu
is anyone able to answer this?
how many moles of Mg do you have?
The equation says you will get that many moles of Mg(NO3)2
But you will really only get 0.3080 times that number of moles
Now convert that to grams.
To calculate the actual yield of magnesium nitrate formed, we need to use the given efficiency and the amount of magnesium used.
First, we need to convert the mass of magnesium (147.4 g) to moles. To do this, we can use the molar mass of magnesium (24.31 g/mol):
Number of moles of magnesium = Mass of magnesium / Molar mass of magnesium
= 147.4 g / 24.31 g/mol
= 6.07 mol
According to the balanced equation: Mg + Cu(NO3)2 → Mg(NO3)2 + Cu, we can see that the stoichiometric ratio between magnesium and magnesium nitrate is 1:1. This means that for every 1 mole of magnesium, we will obtain 1 mole of magnesium nitrate.
Since the efficiency of the reaction is given as 30.80%, we can calculate the theoretical yield by multiplying the number of moles of magnesium by the efficiency:
Theoretical yield = Number of moles of magnesium x Efficiency
= 6.07 mol x 0.3080
= 1.87 mol
However, it is mentioned that copper(II) nitrate is in excess, which means there will be no limitation on the formation of magnesium nitrate. Thus, the actual yield will be equal to the theoretical yield.
Therefore, the actual yield of magnesium nitrate formed is 1.87 mol.