\(PTHH:2A+3H_2SO_4--->A_2\left(SO_4\right)_3+3H_2\left(1\right)\)
Ta có: \(n_{H_2}=\dfrac{\dfrac{672}{1000}}{22,4}=0,03\left(mol\right)\)
a. Theo PT(1): \(n_A=\dfrac{2}{3}.n_{H_2}=\dfrac{2}{3}.0,03=0,02\left(mol\right)\)
\(\Rightarrow M_A=\dfrac{0,54}{0,02}=27\left(\dfrac{g}{mol}\right)\)
Vậy A là nguyên tố nhôm (Al)
b. \(PTHH:2Al+3H_2SO_4--->Al_2\left(SO_4\right)_3+3H_2\left(2\right)\)
Theo PT(2): \(n_{H_2SO_4}=n_{H_2}=0,03\left(mol\right)\)
\(\Rightarrow C_{\%_{H_2SO_4}}=\dfrac{0,03.98}{100}.100\%=2,94\%\%\)
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