When 15 grams of copper (ii) chloride (cucl2) reacts with 20 grams of sodium nitrate (nano3), 11.3 grams of sodium chloride (nacl) is obtained. what is the percent yield of this reaction?
cucl2 + 2nano3 cu(no3)2 + 2nacl

Answer : The percent yield of the reaction is, 86.3 %

Solution : Given,

Mass of = 15 g

Mass of = 20 g

Molar mass of = 134.45 g/mole

Molar mass of = 84.9 g/mole

Molar mass of = 58.44 g/mole

First we have to calculate the moles of and .

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

From the balanced reaction we conclude that

As, 1 mole of react with 2 mole of

So, 0.112 moles of react with moles of

From this we conclude that, is an excess reagent because the given moles are greater than the required moles and is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of

From the reaction, we conclude that

As, 1 mole of react to give 2 mole of

So, 0.112 moles of react to give moles of

Now we have to calculate the mass of

Theoretical yield of = 13.09 g

Experimental yield of = 11.3 g

Now we have to calculate the percent yield of

Therefore, the percent yield of the reaction is, 86.3 %

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4d^55s^1 ( third choice)