Consider the reaction represented by the equation. 2NaHCO3(s) \(\rightarrow\) 2Na2CO 3(S) + CO 2(g) + H2O(g) what volume of carbon (IV)oxide at s.t.p is evolved when 0. 5 moles of NaHCO3 is heated? [Molar volume = 22.4 dm3 at s.t. p]
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Correct Answer: Option C
Explanation:
2NaHCO3(s) \(\rightarrow\) 2Na2CO 3(S) + CO 2(g) + H2O(g)
2 moles of 2NaHCO3(s) produces 22.4dm3 of CO2
0.5 moles of 2NaHCO3 will produce \(\frac{0.5 \times 22.4}{2}\)
= \(\frac{11.2}{2}\)
= 5.6 dm3
2NaHCO3(s) \(\rightarrow\) 2Na2CO 3(S) + CO 2(g) + H2O(g)
2 moles of 2NaHCO3(s) produces 22.4dm3 of CO2
0.5 moles of 2NaHCO3 will produce \(\frac{0.5 \times 22.4}{2}\)
= \(\frac{11.2}{2}\)
= 5.6 dm3