Question

Question #55520

Answer 1

33.6L of `CO_2`

Explanation:

First write the equation

`2NaHCO_3 rarr Na2CO3(s) + CO2(g) + H2O(l)`

The mole ratio of reactant : product`(CO_2)`
= 2mol : 1mol
Solve for ratio
Amount of CO2 formed in moles =

3mol : x = 2mol : 1mol

3 and x should be in the proportion so that `3/x = 2/1`

Remember
products of means = product of extremes

`color(red)(3) : color(blue)(x) = color(blue)(2) : color (red)(1)`

Or `2x = 3xx1`

x = 1.5mol

Volume of 1mol at STP = 22.4L
Thus volume of 1.5mol at STP

`= 1.5mol xx 22.4L = 33.6L`