For question 3 and question 4, we have:
a) The mass of aluminum sulfate in a sample that has 1.14 moles of sulfate anions is 130 g.
b) The mass of copper(II) sulfate pentahydrate in a sample that has 6.62 g of copper is 26.01 g.
a) The chemical formula of aluminum sulfate is Al₂(SO₄)₃.
So, in 1 mol of Al₂(SO₄)₃ we have 2 moles of Al³⁺ (aluminum cations) and 3 moles of SO₄²⁻ (sulfate anions). If we have 1.14 moles of sulfate anions, then the number of moles of aluminum sulfate is:
[tex]n_{Al_{2}(SO_{4})_{3}} = \frac{1\: mol_{Al_{2}(SO_{4}^{2-})_{3}}}{3 \:moles_{SO_{4}^{2-}}}*1.14 \:moles_{SO_{4}^{2-}} = 0.38 \:moles[/tex]
Now, we can find the mass of Al₂(SO₄)₃:
[tex] m_{Al_{2}(SO_{4})_{3}} = n_{Al_{2}(SO_{4})_{3}}*M = 0.38 moles*342.15 g/mol= 130 g [/tex]
Hence, the mass of aluminum sulfate is 130 g.
b) The chemical formula of copper(II) sulfate pentahydrate is CuSO₄·5H₂O.
Since in 1 mol of CuSO₄·5H₂O we have 1 mol of Cu²⁺, the mass of CuSO₄·5H₂O can be calculated as follows:
[tex] m_{CuSO_{4}\cdot 5H_{2}O} = \frac{1 \:mol_{CuSO_{4}\cdot 5H_{2}O}}{1 \: mol_{Cu}}*\frac{1 \: mol_{Cu}}{63.55 g_{Cu}}*\frac{249.68 g_{CuSO_{4}\cdot 5H_{2}O}}{1 \:mol_{CuSO_{4}\cdot 5H_{2}O}}*6.62 \:g_{Cu} = 26.01 g [/tex]
Therefore, the mass of copper(II) sulfate pentahydrate is 26.01 g.
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