\(a,V_{SO_3}=n\cdot22,4=\dfrac{4}{32+16\cdot3}\cdot22,4=1,12\left(l\right)\\ b,V_{CO_2}=n\cdot22,4=\dfrac{22}{12+16\cdot2}\cdot22,4=11,2\left(l\right)\\ c,n_{H_2}=\dfrac{12\cdot10^{-23}}{6\cdot10^{-23}}=2\left(mol\right)\\ \Rightarrow V_{H_2}=2\cdot22,4=44,8\left(l\right)\\ d,V_{N_2}=0,025\cdot22,4=0,56\left(l\right)\)
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