Question

For the following reaction, 4.57 g of silver nitrate are mixed with excess copper. The reaction yields 2.29 gram of copper(II) nitrate What is the percent yield for this reaction?Formula: % yield = (Actual yield/theoretical yield) x 100 2 AgNO3(aq) + Cu(s) à Cu(NO3)2 (aq) + 2 Ag(s)

Answer 1

`Y=90.6\%`

Step-by-step explanation:

Hello,

In this case, for the given chemical reaction:

`2AgNO_3(aq) + Cu(s) \rightarrow Cu(NO_3)_2 (aq) + 2 Ag(s)`

If 4.57 grams of silver nitrate are used in copper excess, the theoretical yield of copper (II) nitrate turns out:

`m_(Cu(NO_3)_2)^(theoretical)=4.57gAgNO_3*(1molAgNO_3)/(170gAgNO_3)*(1molCu(NO_3)_2)/(2molAgNO_3)*(188gCu(NO_3)_2)/(1molCu(NO_3)_2) \\\\m_(Cu(NO_3)_2)^(theoretical)=2.53gCu(NO_3)_2`

In such a way, the percent yield results:

`Y=(2.29g)/(2.53g)*100\%=90.6\%`

Best regards.