Question

94g NaHCO_(3)*(1 mole NaHCO_(3))/(84g NaHCO_(3))*(1 mole NaCl)/(1 mole NaHCO_(3))*(58.5g NaCl)/(1 mole NaCl)

Answer 1

$\begin{aligned} &94g NaHCO_3 × \dfrac{1\ mole \ NaHCO_3}{84g\ NaHCO_3}×\\ \\ &\dfrac{1\ mole \ NaCl}{1\ mole\ NaHCO_3} × \dfrac{58.5g \ NaCl}{1\ mole\ NaCl} = \end{aligned}$

Re-arranging the numerators and denominations, so that$NaHCO_3$ and $NaCl$ are grouped together, and moles and masses are also grouped together, we have;

$\begin{aligned} &\dfrac{1\ mole\ of\ NaHCO_3}{1\ mole\ NaHCO_3} × \dfrac{94g NaHCO_3 }{84g\ NaHCO_3}×\\ \\ &\dfrac{1\ mole \ NaCl}{1\ mole\ NaCl} × 58.5g \ NaCl= \end{aligned}$

$\begin{aligned} 1 ×1.12 × 1 ×58.5g \ NaCl= \end{aligned}$

$65.5g\ of\ NaCl$