Answer: The mass of Cu produced is 4.88 g
Explanation:
The number of moles is defined as the ratio of the mass of a substance to its molar mass.
The equation used is:
\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}} ……(1)
Given mass of aluminum = 2.98 g
Molar mass of aluminum = 27 g/mol
Plugging values in equation 1:
\text{Moles of aluminum}=\frac{2.98g}{27g/mol}=0.1104 mol
The given chemical equation follows:
2Al(s)+3CuSO_4(aq)\rightarrow Al_2(SO_4)_3(aq)+3Cu(s)
By the stoichiometry of the reaction:
If 2 moles of aluminum produces 3 moles of Cu
So, 0.1104 moles aluminium will produce = \frac{3}{2}\times 0.1104=0.1656mol of Cu
Molar mass of Cu = 63.5 g/mol
Plugging values in equation 1:
\text{Mass of Cu}=(0.1656mol\times 63.5g/mol)=10.516g
The percent yield of a reaction is calculated by using an equation:
\% \text{yield}=\frac{\text{Actual value}}{\text{Theoretical value}}\times 100 ……(2)
Given values:
% yield of product = 46.4 %
Theoretical value of the product = 10.516 g
Plugging values in equation 2, we get:
46.4=\frac{\text{Actual value of Cu}}{10.516g}\times 100\\\\\text{Actual value of Cu}=\frac{46.4\times 10.516}{100}\\\\\text{Actual value of Cu}=4.88g
Hence, the mass of Cu produced is 4.88 g
Answer:
-753kJ of energy are involved
Explanation:
Based on the reaction:
Fe₂O₃(s) + 2 Al(s) → Al₂O₃(s) + 2 Fe(s) ΔH°rxn = -852 kJ
<em>When 2 moles of Fe are produced, there are released -852kJ.</em>
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98.7g of Fe are:
98.7g Fe × (1mol / 55.845g) = 1.767 moles of Fe
If 2 moles of Fe are producen when -852 kJ of energy are involved, 1.767 moles of Fe envolved:
1.767mol × (-852kJ / 2mol Fe) = <em>-753kJ of energy are involved</em>
Explanation:
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