\(2.m_{NaCl}=1.58,5=58,5\left(g\right)\\ m_{H_2O}=3.18=54\left(g\right)\\ Vì:58,5>54\Leftrightarrow m_{NaCl}>m_{H_2O}\\ 3.n_{O_2}=\dfrac{4,8}{32}=0,15\left(mol\right)\\ V_{O_2\left(đktc\right)}=0,15.22,4=3,36\left(l\right)\\ 4.n_{O_2}=\dfrac{64}{32}=2\left(mol\right)\\ n_{SO_2}=\dfrac{64}{64}=1\left(mol\right)\)
Vì: n tỉ lệ thuận với V
=> \(V_{O_2}>V_{SO_2}\)
\(5.n_{CO_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\\ m_{CO_2}=44.0,25=11\left(g\right)\)
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