Question

Answer 1

a) \(n_{Cu}=\dfrac{N}{A}=\dfrac{9,033.10^{23}}{6.10^{23}}=1,5\left(mol\right)\\ m_{Cu}=1,5.64=96\left(g\right)\)

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\(n_{NO_2}=\dfrac{N}{A}=\dfrac{4,5165.10^{23}}{6.10^{23}}=0,75\left(mol\right)\\ m_{NO_2}=0,75.46=34,5\left(g\right)\)

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b) \(n_{H_2S}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\\ m_{H_2S}=0,6.34=20,4\left(g\right)\)

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\(n_{N_2}=\dfrac{33,6}{22,4}=1,5\left(mol\right)\\ m_{N_2}=1,5.28=42\left(g\right)\)

 

 

Answer 2

a) n_Cu= \(\dfrac{9,033.10^{23}}{6.10^{23}}\)= 1,5055 mol => m_Cu=1,5055.64=96,352 g

b) n_H2S= \(\dfrac{13,44}{22,4}\)= 0,6 mol   => m_H2S= 0,6.34= 20,4 g