rachie19
  • rachie19
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
Chemistry
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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rachie19
  • rachie19
@Rushwr
Rushwr
  • Rushwr
Do u know how to get this ?
Rushwr
  • Rushwr
Efficiency = actual yield/ theoretical yield *100%

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Rushwr
  • Rushwr
First we will have to find the theoretical yield first !
rachie19
  • rachie19
yh i got 5.841217606g for my actual yiel
Rushwr
  • Rushwr
oh let me check
rachie19
  • rachie19
but im stuck
Rushwr
  • Rushwr
where are u sucked?
rachie19
  • rachie19
im not sure what to do next, im trying to find the theoretical yield
rachie19
  • rachie19
i mean actual
rachie19
  • rachie19
the one i did was the theoretical i wrote actual
Rushwr
  • Rushwr
oh still wrong ! Wait. I'll tell u .
rachie19
  • rachie19
ok thanks
Rushwr
  • Rushwr
First u have to find a relationship between Mg and Mg(NO3)2 ! Cuz the information we know is about Mg right?
Photon336
  • Photon336
\(\color{blue}{\text{Originally Posted by}}\) @Rushwr First we will have to find the theoretical yield first ! \(\color{blue}{\text{End of Quote}}\) \[\frac{ actual }{ theoretical}*100 = percent, yield \]
rachie19
  • rachie19
to get the theoretical yield. i divided the grams which was given.... 142.4/ 24.31 (1mol (NO3)2/ 1mol of Mg
rachie19
  • rachie19
i then multiplied by 148.31 g/mol
Rushwr
  • Rushwr
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
Rushwr
  • Rushwr
I hope u got it @rachie19 :)
rachie19
  • rachie19
ok i understand what i did wrong. Thank you so much :)
Rushwr
  • Rushwr
Ur welcome !!! And no problem !

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