A sample of gas contains 0.1600 mol of C_{2}H_{4}(g) and 0.1600 mol of Br_{2}(g) and occupies a volume of 10.4 L. The following reaction takes place:

Solution:

According with Avogadro's Law (The Volume Amount Law)

(V_{1} / n_{1} = V_{2} / n_{2})

before the reaction takes place the total number of moles is = 0.16+0.16= 0.32 moles

and the volume is 10.4 L

According to the balanced chemical equation, 1 mol of C2H4(g) produces 1 mol of CH2BrCH2Br(g)

then 0.16 mol of C2H4(g) produces 0.16 mol of CH2BrCH2Br(g)

then the number of moles after the reaction is 0.16

Now we can solve the avogadro´s law for V2

V2=V1n2/n1

V2= (10.4L)*(0.16mol)/(0.32mol)= 5.2 L

The Volume after the reaction is 5.2 L

can you do this one for me please!

A sample of gas contains 0.1200 mol of OF_{2}(g) and 0.1200 mol of H_{2}O(g) and occupies a volume of 7.92 L. The following reaction takes place:OF_{2}(g) + H_{2}O(g)O_{2}(g) + 2HF(g)Calculate the volume of the sample after the reaction takes place, assuming that the temperature and the pressure remain constant. L

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You can put an easy question (I think is free) and I´ll answer your question

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