Assuming an efficiency of 29.40%, how would you calculate the actual yield of magnesium nitrate formed from 123.5 g of magnesium and excess copper(II) nitrate in the equation #Mg + Cu(NO_3)_2 -> Mg(NO_3)_2 + Cu#?

Answer 1

#"221.5 g Mg"("NO"_ 3)_2#

The first thing to do here is to calculate the theoretical yield of the reaction, i.e. what you'd expect to see for a reaction that has a #100%# yield.

Take note of how that affects your response.

#"Mg"_ ((s)) + "Cu"("NO"_ 3)_ (2(aq)) -> "Mg"("NO"_ 3)_ (2(aq)) + "Cu"_ ((s))#

every #1# mole of magnesium metal that takes part in the reaction consumes #1# mole of copper(II) nitrate and produces #1# mole of aqueous magnesium nitrate.

Use the molar mass of the element to convert the mass of magnesium to moles.

#123.5 color(red)(cancel(color(black)("g"))) * "1 mole Mg"/(24.305color(red)(cancel(color(black)("g")))) = "5.081 moles Mg"#

Since there is an excess of copper(II) nitrate, you can presume that every mole of magnesium participates in the reaction.

This means that at #100%# yield, you would get

#5.081 color(red)(cancel(color(black)("moles Mg"))) * ("1 mole Mg"("NO"_ 3)_ 2)/(1color(red)(cancel(color(black)("mole Mg")))) = "5.081 moles Mg"("NO"_ 3)_2#

Now, you know that the reaction has a #29.40%# yield, which means that for every #100# moles of magnesium nitrate that could be produced, only #29.40# moles are actually produced.

Consequently, this response will result in

#5.081 color(red)(cancel(color(black)("moles Mg"("NO"_ 3)_ 2))) * ("29.40 moles Mg"("NO"_ 3)_ 2)/(100color(red)(cancel(color(black)("moles Mg"("NO"_ 3)_ 2))))#

# = "1.4938 moles Mg"("NO"_ 3)_ 2#

Use the molar mass of the compound to convert this to grams.

#1.4938 color(red)(cancel(color(black)("moles Mg"("NO"_ 3)_ 2))) * "148.3 g"/(1color(red)(cancel(color(black)("mole Mg"("NO"_ 3)_ 2)))) = color(darkgreen)(ul(color(black)("221.5 g")))#

Four sig figs are used to round the result.

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Answer 2

Cameron Andrews

To calculate the actual yield of magnesium nitrate formed, we first need to determine the theoretical yield of magnesium nitrate using stoichiometry based on the given amount of magnesium. Then, we can use the efficiency percentage to find the actual yield.

Calculate the molar mass of magnesium nitrate (Mg(NO3)2): Mg: 1 atom (24.305 g/mol) N: 2 atoms (14.007 g/mol) O: 6 atoms (16.00 g/mol) Total molar mass = (1 * 24.305) + (2 * 14.007) + (6 * 16.00) = 148.32 g/mol
Determine the theoretical yield of magnesium nitrate: Using stoichiometry, 1 mole of magnesium (24.305 g) reacts with 1 mole of copper(II) nitrate (Cu(NO3)2) to produce 1 mole of magnesium nitrate (Mg(NO3)2). Given 123.5 g of magnesium, convert it to moles: Moles of magnesium = mass / molar mass = 123.5 g / 24.305 g/mol Then, the moles of magnesium nitrate produced will be the same as the moles of magnesium used.
Calculate the theoretical yield of magnesium nitrate: Theoretical yield = moles of magnesium * molar mass of Mg(NO3)2
Once the theoretical yield is calculated, multiply it by the efficiency percentage to find the actual yield of magnesium nitrate formed.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

Assuming an efficiency of 29.40%, how would you calculate the actual yield of magnesium nitrate formed from 123.5 g of magnesium and excess copper(II) nitrate in the equation #Mg + Cu(NO_3)_2 -&gt; Mg(NO_3)_2 + Cu#?

Assuming an efficiency of 29.40%, how would you calculate the actual yield of magnesium nitrate formed from 123.5 g of magnesium and excess copper(II) nitrate in the equation #Mg + Cu(NO_3)_2 -> Mg(NO_3)_2 + Cu#?