\(n_{NaHCO_3}=\frac{84}{84}=1\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{Na}=1\left(mol\right)\\n_H=1\left(mol\right)\\n_C=1\left(mol\right)\\n_O=3\left(mol\right)\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m_{Na}=23\left(g\right)\\m_H=1\left(g\right)\\m_C=12\left(g\right)\\m_O=48\left(g\right)\end{matrix}\right.\)