Answer
a. $2NaHCO_{3}+O_{2}$ -> $2CO_{2}+Na_{2}CO_{3}+H_{2}O$
b. $39.1$ $g$ $NaHCO_{3}$
Work Step by Step
a. $2NaHCO_{3}+O_{2}$ -> $2CO_{2}+Na_{2}CO_{3}+H_{2}O$ (BALANCE THE EQUATION)
b. Mass of $NaGCO_{3}$ =
$\frac{(20.5gCO_{2})\times(1molCO_{2})\times(2molNaHCO_{3})\times(84.01gNaHCO_{3})}{(44.01gCO_{2})\times(2molCO_{2})\times(1molNaHCO_{3})}$
$\approx$39.13217...
= $39.1$ $g$ $NaHCO_{3}$
