Final answer:
To calculate the actual yield of magnesium nitrate from 117.8 g of magnesium with a 49.10% efficiency, we first find the theoretical yield through stoichiometry, then multiply by the efficiency. The actual yield is the product of the efficiency and theoretical yield. Note that the reference information incorrectly mentions Zn(NO3)2 instead of Mg(NO3)2, but the calculation principles remain the same.
Step-by-step explanation:
To calculate the actual yield of magnesium nitrate formed from 117.8 g of magnesium, we first need to determine the theoretical yield. To do this, we start by finding the molar masses of the reactants and products involved in the reaction: Mg(s) + Cu(NO3)2 (aq) → Mg(NO3)2 (aq) + Cu(s). The molar mass of Mg is 24.31 g/mol, and the molar mass of Mg(NO3)2 is 148.31 g/mol. Using stoichiometry, we can set up the mass-mass calculation to find the theoretical yield.
Once the theoretical yield is calculated, we can then use the given efficiency of 49.10% to calculate the actual yield. The actual yield is the product of the efficiency (expressed as a decimal) and the theoretical yield. To get the percentage yield, we would divide the actual yield by the theoretical yield and multiply by 100.
It appears there is a mistake in the provided reference information, as it refers to the substance Zn(NO3)2 instead of Mg(NO3)2. Nonetheless, we can apply the same principles to calculate the yield for magnesium nitrate from the starting mass of magnesium.