Amount of substance

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Question 3
Easy

The equation for the complete combustion of gaseous methylamine is:

4CH3NH2(g)+9O2(g)→4CO2(g)+10H2O(g)+2N2(g) 4\text{CH}_3\text{NH}_2(\text{g}) + 9\text{O}_2(\text{g}) \rightarrow 4\text{CO}_2(\text{g}) + 10\text{H}_2\text{O}(\text{g}) + 2\text{N}_2(\text{g}) 4CH3​NH2​(g)+9O2​(g)→4CO2​(g)+10H2​O(g)+2N2​(g)

What is the mole fraction of methylamine in a gaseous reactant mixture containing the minimum quantity of oxygen required for complete combustion?

0.1380.1380.138

0.3080.3080.308

0.4440.4440.444

0.6920.6920.692

Amount of substance Questions

  1. A Level
  2. /Chemistry
  3. /Amount of substance