Answer to Question #95856 in General Chemistry for Kyli Dean

Question #95856
Assuming an efficiency of 31.00%, calculate the actual yield of magnesium nitrate formed from 127.3 g of magnesium and excess copper(II) nitrate.

Mg + Cu(NO3)2 ⟶ Mg(NO3)2 + Cu
1
Expert's answer
2019-10-08T07:40:26-0400

Moles of "Mg=\\frac{127.3}{24}=5.30mol"

So according to the given efficiency moles of "Mg(NO_3)_2" formed"=" "0.31\\times5.30=1.643mol"

Hence mass of "Mg(NO_3)_2" formed"=1.643\\times148=243.164g"


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