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Sagot :
Answer: 21.13 L of [tex]H_2O[/tex] will be produced from the reaction of 24.65 L of oxygen
Explanation:
According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number [tex]6.023\times 10^{23}[/tex] of particles.
[tex]2C_2H_6(g)+7O_2(g)\rightarrow 4CO_2(g)+6H_2O(g)[/tex]
According to stoichiometry :
7 moles of [tex]O_2[/tex] give = 6 moles of [tex]H_2O[/tex] vapour
[tex]7\times 22.4L=156.8L[/tex] of [tex]O_2[/tex] produce = [tex]6\times 22.4L=134.4L[/tex] of [tex]H_2O[/tex] vapour
Thus 24.65 L of [tex]O_2[/tex] produce = [tex]\frac{134.4}{156.8}\times 24.65=21.13[/tex] L of [tex]H_2O[/tex] vapour
21.13 L of [tex]H_2O[/tex] will be produced from the reaction of 24.65 L of oxygen
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