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. write the number in grams.
Mg+Cu(NO3)2=Mg(NO3)2+Cu
is anyone able to solve this? ive gotten the answer wrong twice
moles Mg = 147.4 g / molar mass Mg
moles Mg(NO3)2 = moles Mg * .3080
mass Mg(NO3)2 = moles Mg(NO3)2 * molar mass Mg(NO3)2
To Skye---What is your problem. You have posted this problem at least three times. Oobleck answered it, I've answered it, R-Scott has answered it. You keep posting. Get a grip.
To calculate the actual yield of magnesium nitrate formed, you need to use the given information about the efficiency and the initial mass of magnesium.
1. Start by converting the mass of magnesium to moles. To do this, use the molar mass of magnesium, which is 24.31 g/mol:
Moles of Mg = Mass of Mg / Molar mass of Mg = 147.4 g / 24.31 g/mol
2. Use the balanced equation and the moles of magnesium to determine the stoichiometry between magnesium and magnesium nitrate. From the balanced equation:
1 mol Mg produces 1 mol Mg(NO3)2
3. Now, convert the moles of magnesium to moles of magnesium nitrate. Since the stoichiometry is 1:1, the moles will remain the same.
4. Next, use the given efficiency of 30.80% to calculate the theoretical yield of magnesium nitrate. The theoretical yield is the maximum amount of magnesium nitrate that could be formed if the reaction were 100% efficient.
Theoretical yield = Moles of Mg(NO3)2 * Molar mass of Mg(NO3)2
5. Finally, calculate the actual yield by multiplying the theoretical yield by the efficiency percentage.
Actual yield = Theoretical yield * Efficiency
Make sure to use accurate values for the molar masses. By following these steps, you should be able to calculate the actual yield of magnesium nitrate formed in grams.