Bài 16
a) \(m_{Cl}=\dfrac{58,5.60,68\%}{100\%}=35\left(g\right)\)
\(n_{Cl}=\dfrac{35}{35,5}=1\left(mol\right)\)
\(m_{Na}=58,5-35=23,5\left(g\right)\\ n_{Na}=\dfrac{23,5}{23}=1\left(mol\right)\)
\(=>CTHH:NaCl\)
b) \(m_{Na}=\dfrac{106.43,4\%}{100\%}=46\left(g\right)\)
\(n_{Na}=\dfrac{46}{23}=2\left(mol\right)\)
\(m_C=\dfrac{106.11,3\%}{100\%}=12\left(g\right)\\ n_C=\dfrac{12}{12}=1\left(mol\right)\)
\(m_O=106-46-12=48\left(g\right)\\ n_O=\dfrac{48}{16}=3\left(mol\right)\\ =>CTHH:Na_2CO_3\)
\(\dfrac{M_m}{M_{H_2}}=8,5\Rightarrow8,5.2=17\left(gmol\right)\)
\(m_N=\dfrac{17.82,35\%}{100\%}=14\left(g\right)\\ n_N=\dfrac{14}{14}=1\left(mol\right)\)
\(m_H=17-14=3\left(g\right)\\ n_H=\dfrac{3}{1}=3\left(mol\right)\)
\(=>CTHH:NH_3\)
