Question

[huy giang nguyễn trần]

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Answer 1

[GV Nguyễn Trần Thành Đạt]

\(a,n_{Cu}=\dfrac{m_{Cu}}{M_{Cu}}=\dfrac{256}{64}=4\left(mol\right)\\ b,n_{FeCl_3}=\dfrac{m_{FeCl_3}}{M_{FeCl_3}}=\dfrac{3,25}{162,5}=0,02\left(mol\right)\\ c,n_{NaOH}=\dfrac{m_{NaOH}}{m_{NaOH}}=\dfrac{6}{40}=0,15\left(mol\right)\\ d,n_{H_2S}=\dfrac{V_{H_2S\left(đkc\right)}}{24,79}=\dfrac{4,958}{24,79}=0,2\left(mol\right)\)

Answer 2

[huy giang nguyễn trần]

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