Question #55520

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Answer 1 · 2017-03-27T14:03:27

33.6L of #CO_2#

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#