(Solved) ACTUAL YIELD

dacia79

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10 years ago

Assuming an efficiency of 28.00%, calculate the actual yield of magnesium nitrate formed from 138.0 g of magnesium and excess copper(II) nitrate. Mg+Cu(NO3)2 -->Mg (NO3)2+Cu

I know that actual yield =percent yield x theoretical yield/100, but I don't know how to get those numbers. Please help. Thank you.

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padre

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10 years ago

Similar Question

Assuming an efficiency of 44.00%, calculate the actual yield of magnesium nitrate formed from 110.7 g of magnesium and excess copper(II) nitrate.
Mg + Cu(NO3)2 Rightwards Arrow

Mg(NO3)2 + Cu

Answer

110.7 g Mg will react with 854.2 g Cu(NO3)2 to form 675.5 g Mg(NO3)2 and 289.4 g Cu.

The actual yield of magnesium nitrate is (675.5 g) x 0.44

Just replace the numbers with the numbers found in your question.

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Ezekiel

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10 years ago

Padre, thank you first and foremost for the help, but I still don't understand where you got the large numbers such as 854, 675 and so on. I understand what needs to be done but not how I can do it. I know that I have to: 1. Convert the mass of Mg to moles. 2. Find the corresponding moles of Mg(NO3)2 using the coefficients in the balanced chemical equation. 3. Convert the moles of Mg(NO3)2 to grams. If you can dumb it down I would appreciate it because my brain is fried. Thank you.

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padre

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10 years ago

Sorry, I don't know. I just found that online. I assumed the person would know how to take it from there.

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