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Determine the amount of copper, in grams, that can be recovered at the end of the last step, starting with 0.90 g of copper in the first step. The percentage yield of copper in each step is [tex]$99 \%$[/tex].

[tex]
\begin{array}{l}
Cu ( s )+4 HNO _3( aq ) \rightarrow Cu \left( NO _3\right)_2( aq )+2 H _2 O ( l )+2 NO _2(g) \\
Cu \left( NO _3\right)_2( aq )+2 NaHCO _3( aq ) \rightarrow CuCO _3(s)+2 NaNO _{3}( aq )+ H _2 O ( l )+ CO _2(g) \\
CuCO _3(s)+ H _2 SO _4( aq ) \rightarrow CuSO _4( aq )+ H _2 O ( l )+ CO _2(g) \\
Zn ( s )+ CuSO _4( aq ) \rightarrow Cu ( s )+ ZnSO _4( aq )
\end{array}
[/tex]

Given initial amount of copper: [tex]$0.90 \, \text{g}$[/tex]


Sagot :

To determine the amount of copper that can be recovered after the final step given a series of chemical transformations, we must account for the percentage yield at each step. Here's the step-by-step solution:

1. Understanding the Initial Conditions:
- We start with 0.90 grams of copper.

2. Percentage Yield per Step:
- The percentage yield of copper in each step is [tex]\( 99\% \)[/tex]. This means we retain [tex]\( 99\% \)[/tex] of the copper after each transformation.

3. Convert the Percentage to a Decimal:
- [tex]\( 99\% \)[/tex] can be written as [tex]\( 0.99 \)[/tex] in decimal form.

4. Number of Steps:
- There are 4 steps in the chemical transformation series.

5. Calculate the Final Amount of Copper:
- In each step, the amount of copper is multiplied by [tex]\( 0.99 \)[/tex] to account for the yield.
- Since there are 4 steps, the final amount of copper [tex]\( A \)[/tex] can be calculated using the formula:
[tex]\[ A = \text{initial amount of copper} \times (\text{yield percentage})^n \][/tex]
where [tex]\( n \)[/tex] is the number of steps.

6. Substitute the Values:
- Initial amount of copper = [tex]\( 0.90 \)[/tex] grams.
- Yield percentage per step = [tex]\( 0.99 \)[/tex].
- Number of steps ([tex]\( n \)[/tex]) = 4.
- Therefore:
[tex]\[ A = 0.90 \times (0.99)^4 \][/tex]

7. Perform the Calculation:
- [tex]\( 0.99^4 \approx 0.96059601 \)[/tex]
- So, [tex]\( A = 0.90 \times 0.96059601 \approx 0.864536409 \)[/tex]

Therefore, the amount of copper that can be recovered at the end of the last step is approximately [tex]\( 0.8645 \)[/tex] grams.