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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
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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Rushwr
 one year ago
Best ResponseYou've already chosen the best response.2Do u know how to get this ?

Rushwr
 one year ago
Best ResponseYou've already chosen the best response.2Efficiency = actual yield/ theoretical yield *100%

Rushwr
 one year ago
Best ResponseYou've already chosen the best response.2First we will have to find the theoretical yield first !

rachie19
 one year ago
Best ResponseYou've already chosen the best response.0yh i got 5.841217606g for my actual yiel

rachie19
 one year ago
Best ResponseYou've already chosen the best response.0im not sure what to do next, im trying to find the theoretical yield

rachie19
 one year ago
Best ResponseYou've already chosen the best response.0the one i did was the theoretical i wrote actual

Rushwr
 one year ago
Best ResponseYou've already chosen the best response.2oh still wrong ! Wait. I'll tell u .

Rushwr
 one year ago
Best ResponseYou've already chosen the best response.2First u have to find a relationship between Mg and Mg(NO3)2 ! Cuz the information we know is about Mg right?

Photon336
 one year ago
Best ResponseYou've already chosen the best response.0\(\color{blue}{\text{Originally Posted by}}\) @Rushwr First we will have to find the theoretical yield first ! \(\color{blue}{\text{End of Quote}}\) \[\frac{ actual }{ theoretical}*100 = percent, yield \]

rachie19
 one year ago
Best ResponseYou've already chosen the best response.0to get the theoretical yield. i divided the grams which was given.... 142.4/ 24.31 (1mol (NO3)2/ 1mol of Mg

rachie19
 one year ago
Best ResponseYou've already chosen the best response.0i then multiplied by 148.31 g/mol

Rushwr
 one year ago
Best ResponseYou've already chosen the best response.2As 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
 one year ago
Best ResponseYou've already chosen the best response.2I hope u got it @rachie19 :)

rachie19
 one year ago
Best ResponseYou've already chosen the best response.0ok i understand what i did wrong. Thank you so much :)

Rushwr
 one year ago
Best ResponseYou've already chosen the best response.2Ur welcome !!! And no problem !
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