12.4 g of copper(II) carbonate are heated and decomposes. 6 g of copper(II) oxide is formed. Calculate the percentage yield.
Answers
Answer : The percentage purity in copper carbonate is, 87.6 %
Explanation : Given,
Mass crude copper carbonate = 12.4 g
Mass of cooper(II) oxide = 7 g
Molar mass of copper carbonate = 123.5 g/mole
Molar mass of copper(II) oxide = 79.5 g/mole
The balanced chemical reaction will be,
CuCO_3\rightarrow CuO+CO_2CuCO
3
→CuO+CO
2
First we have to calculate the moles of CuO.
\text{Moles of }CuO=\frac{\text{Mass of }CuO}{\text{Molar mass of }CuO}=\frac{7g}{79.5g/mole}=0.088molesMoles of CuO=
Molar mass of CuO
Mass of CuO
=
79.5g/mole
7g
=0.088moles
Now we have to calculate the moles of CuCO_3CuCO
3
From the balanced reaction we conclude that
As, 1 mole of CuO produced from 1 mole of CuCO_3CuCO
3
So, 0.088 mole of CuO produced from 0.088 mole of CuCO_3CuCO
3
Now we have to calculate the mass of CuCO_3CuCO
3
.
\text{Mass of }CuCO_3=\text{Moles of }CuCO_3\times \text{Molar mass of }CuCO_3=(0.088mole)\times (123.5g/mole)=10.868gMass of CuCO
3
=Moles of CuCO
3
×Molar mass of CuCO
3
=(0.088mole)×(123.5g/mole)=10.868g
The pure mass of CuCO_3CuCO
3
= 10.868 g
Now we have to calculate the percentage purity in copper carbonate.
\%\text{ Purity in }CuCO_3=\frac{\text{Mass of pure }CuCO_3}{\text{Mass of crude }CuCO_3}\times 100=\frac{10.868g}{12.4g}\times 100=87.6\%% Purity in CuCO
3
=
Mass of crude CuCO
3
Mass of pure CuCO
3
×100=
12.4g
10.868g
×100=87.6%
Therefore, the percentage purity in copper carbonate is, 87.6 %
Answer:
the answer is
Explanation:
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