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Answer 1 · 2017-03-27T14:03:27

33.6L of $C O_{2}$ $CO_2$

First write the equation

$2 N a H C O_{3} \to N a 2 C O 3 (s) + C O 2 (g) + H 2 O (l)$ $2NaHCO_3 rarr Na2CO3(s) + CO2(g) + H2O(l)$

The mole ratio of reactant : product $(C O_{2})$ $(CO_2)$
= 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 $\frac{3}{x} = \frac{2}{1}$ $3/x = 2/1$

Remember
products of means = product of extremes

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

Or $2 x = 3 \times 1$ $2x = 3xx1$

x = 1.5mol

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

$= 1.5 m o l \times 22.4 L = 33.6 L$ $= 1.5mol xx 22.4L = 33.6L$