Question

written response
use the following information to answer the next two questions.
a piece of copper is dropped into a silver nitrate solution with the following results:
initial mass of copper = 17.65 g
final mass of copper = 11.02 g
written response

1. calculate the mass of silver produced.

your answer:
written response

1. if a student performed this experiment and produced 20.3 g of silver metal.

calculate the % yield and percent error for this experiment.

Explanation:

Step1: Calculate mass of copper reacted

The mass of copper reacted is the initial mass minus the final mass. Let $m_{initial}$ be the initial mass of copper and $m_{final}$ be the final mass of copper.
$m_{reacted}=m_{initial}-m_{final}=17.65 - 11.02=6.63$ g

Step2: Write the chemical reaction

The reaction between copper and silver - nitrate is $Cu + 2AgNO_3
ightarrow Cu(NO_3)_2+2Ag$. The molar - ratio of $Cu$ to $Ag$ is $1:2$. The molar mass of $Cu$ is $M_{Cu}=63.55$ g/mol and the molar mass of $Ag$ is $M_{Ag}=107.87$ g/mol.
First, calculate the number of moles of $Cu$ reacted: $n_{Cu}=\frac{m_{reacted}}{M_{Cu}}=\frac{6.63}{63.55}\approx0.1043$ mol.
Then, using the molar - ratio, the number of moles of $Ag$ produced is $n_{Ag}=2n_{Cu}=2\times0.1043 = 0.2086$ mol.
The theoretical mass of $Ag$ produced is $m_{Ag}=n_{Ag}\times M_{Ag}=0.2086\times107.87\approx22.50$ g.

Step3: Calculate the percent yield

The percent yield ($\%$ yield) is given by the formula $\%$ yield$=\frac{actual\ mass}{theoretical\ mass}\times100\%$. The actual mass of $Ag$ produced is $m_{actual}=20.3$ g.
$\%$ yield$=\frac{20.3}{22.50}\times100\%\approx90.22\%$

Step4: Calculate the percent error

The percent error ($\%$ error) is given by the formula $\%$ error$=\frac{|theoretical\ value - actual\ value|}{theoretical\ value}\times100\%$.
$\%$ error$=\frac{|22.50 - 20.3|}{22.50}\times100\%=\frac{2.2}{22.50}\times100\%\approx9.78\%$

Answer:

1. $22.50$ g
2. $\%$ yield: $90.22\%$, $\%$ error: $9.78\%$