Question #50852
1 Answer
Explanation:
Start by writing the balanced chemical equation
#"Mg"_text((s]) + "Cu"("NO"_3)_text(2(aq]) -> "Mg"("NO"_3)_text(2(aq]) + "Cu"_text((s])#
Notice that you have
Notice that you're told that copper(II) nitrate is in excess. This means that you have more copper(II) nitrate than you need for the reaction.
Simply put, when one reactant is in excess, you know for a fact that all the moles of the second reactant will take part in the reaction.
Use magnesium's molar mass to determine how many moles of magnesium metal will take part in the reaction
#113.8color(red)(cancel(color(black)("g"))) * "1 mole Mg"/(24.305color(red)(cancel(color(black)("g")))) = "4.682 moles Mg"#
So, 4.682 moles of magnesium will take part in reaction. How many moles of magnesium nitrate would you expect the reaction to produce?
SInce you have a
#4.682color(red)(cancel(color(black)("moles Mg"))) * ("1 mole Mg"("NO"_3)_2)/(1color(red)(cancel(color(black)("mole Mg")))) = "4.682 moles Mg"#
The important thing to realize now is that this is the reaction's theoretical yield.
This is how many moles of magnesium nitrate will be formed when the eaction's percent yield is equal to
However, you know that the reaction's percent yield is actually
This implies that the reaction will not produce as many moles of magnesium nitrate as it would have produced if the percent yield was
To get the actual yield, use the definition of the percent yield, which is
#"% yield" = "actual yield"/"theoretical yield" xx 100#
The actual yield will be equal to
#"actual yield" = ("% yield" xx "theoretical yield")/100#
#"actual yield" = (42.40 * "4.682 moles")/100 = "1.985 moles"#
So, the reaction will only produce 1.985 moles of magnesium nitrate. To get how many grams of the compound you'd get for this many moles, use its molar mass
#1.985color(red)(cancel(color(black)("moles"))) * "148.315 g"/(1color(red)(cancel(color(black)("mole")))) = color(green)("294.4 g Mg"("NO"_3)_2)#
