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rachie19

  • one year ago

Assuming an efficiency of 31.50 % calculate the actual yield of magnesium nitrate formed from 142.4 of magnesium and excess copper (II) nitrate Mg + Cu(NO3)2 = Mg(NO3)2 +Cu

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  1. rachie19
    • one year ago
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    @Rushwr

  2. Rushwr
    • one year ago
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    Do u know how to get this ?

  3. Rushwr
    • one year ago
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    Efficiency = actual yield/ theoretical yield *100%

  4. Rushwr
    • one year ago
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    First we will have to find the theoretical yield first !

  5. rachie19
    • one year ago
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    yh i got 5.841217606g for my actual yiel

  6. Rushwr
    • one year ago
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    oh let me check

  7. rachie19
    • one year ago
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    but im stuck

  8. Rushwr
    • one year ago
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    where are u sucked?

  9. rachie19
    • one year ago
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    im not sure what to do next, im trying to find the theoretical yield

  10. rachie19
    • one year ago
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    i mean actual

  11. rachie19
    • one year ago
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    the one i did was the theoretical i wrote actual

  12. Rushwr
    • one year ago
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    oh still wrong ! Wait. I'll tell u .

  13. rachie19
    • one year ago
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    ok thanks

  14. Rushwr
    • one year ago
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    First u have to find a relationship between Mg and Mg(NO3)2 ! Cuz the information we know is about Mg right?

  15. Photon336
    • one year ago
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    \(\color{blue}{\text{Originally Posted by}}\) @Rushwr First we will have to find the theoretical yield first ! \(\color{blue}{\text{End of Quote}}\) \[\frac{ actual }{ theoretical}*100 = percent, yield \]

  16. rachie19
    • one year ago
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    to get the theoretical yield. i divided the grams which was given.... 142.4/ 24.31 (1mol (NO3)2/ 1mol of Mg

  17. rachie19
    • one year ago
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    i then multiplied by 148.31 g/mol

  18. Rushwr
    • one year ago
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    As we can see in the equation the stoichiometric coefficients are 1:1 in Mg:Mg(NO3)3 That means 1 mole of Mg will form Mg(NO3)2 So the next we are doing is finding the moles of Mg present. we know moles = mass divided by molar mass \[n _{Mg} = \frac{ 142.4g }{ 24gmol ^{-1} }\] Now we are gonna write the same equation for Mg(NO3)2 But here we don't know the mass, so we take the unknown mass as "x" \[n _{Mg(NO3)_{2}} = \frac{ Xg }{ 148.3gmol ^{-1} }\] As I told u earlier moles of Mg = moles of Mg(NO3)2 So we can equalize those 2 equation and find the theoretical mass of Mg(NO3)2 formed. \[\frac{ 142.4 g}{ 24gmol ^{-1} } = \frac{ X g}{ 148.3gmol ^{-1} }\] \[X= 880g \] Now this is the theoretical mass, but we have to find the actual yield . \[efficiency = \frac{ actual yield }{ Theoretical yield } * 100%\] \[actual yield = \frac{ Efficiency* theoretical yield }{ 100 } \] \[Actual yield = \frac{ 31.5 * 880 }{ 100} = 277.2g\] This is the answer I got. U better check my calculations. Others seems okai

  19. Rushwr
    • one year ago
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    I hope u got it @rachie19 :)

  20. rachie19
    • one year ago
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    ok i understand what i did wrong. Thank you so much :)

  21. Rushwr
    • one year ago
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    Ur welcome !!! And no problem !

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