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Assuming an efficiency of 38.60%, calculate the actual yield of magnesium nitrate formed from 145.9 g of magnesium and excess copper(II) nitrate.
I got 230.5g but it isn't right..
I got 6 mols of Mg then calculated molar mass of Mg(NO3)2 and got 148.31 and then multiplied the two together
Then I used the percent yield given to find the actual yield.

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Clare

  1. Mg + Cu(NO3)2 ==> Mg(NO3)2 + Cu
    mols Mg = 145.9/24.3 = 6.003
    The equation is 1 mol to 1 mol; therefore, you will have 6.003 mol Mg(NO3)2 formed. Molar mass I have is 148.31 to give g Mg(NO3)2 =890.29g at 100% yield.
    890.29 x 0.3860 = 343.65g actual yield. You're allowed 4 s.f. and I would round to 343.6g.

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    DrBob222

  2. Thank you!

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    Clare

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