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.

Question

Mg+Cu(NO3)2=Mg(NO3)2+Cu
is anyone able to solve this? ive gotten the answer wrong twice

Answer 1

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

Answer 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.