The mass and Volume is different even thought they have a similar shape and size
Use PV = mRT/M and solve for R. R = PVM/RT. Since you have the same gas under two sets of conditions then you can write
<span>P1V1M1/m1T1 = P2V2M2/m2T2 </span>
<span>Since P, M and T are constant, the equation becomes </span>
<span>V1/m1 = V2/m2 </span>
<span>Now plug in your values and solve for V2</span>
Almost all hydrocarbon 'burn' reactions involve oxygen; it's by far the most reactive substance in air.
<span>Hydrocarbon combustions always involve </span>
<span>[some hydrocarbon] + oxygen --> carbon dioxide + steam. </span>
C6H6(l) + O2 (g)--> CO2 (g)+ H2O (g)
<span>Balance carbon, six on each side: </span>
C6H6(l) + O2 (g)--> 6CO2 (g)+ H2O (g)
<span>Balance hydrogen, six on each side: </span>
C6H6(l) + O2 (g)--> 6CO2(g) + 3H2O (g)
<span>Now, we have fifteen oxygens on the right and O2 on the left. </span>
<span>Two ways to deal with that. We can use a fraction: </span>
C6H6 (l)+ (15/2)O2 (g)--> 6CO2 (g)+ 3H2O (g)
<span>Or, if you prefer to have whole number coefficients, double everything </span>
<span>to get rid of the fraction: </span>
2C6H6 (l)+ 15O2 (g)--> 12CO2 (g)+ 6H2O (g)
<span>With the SATP states thrown in... </span>
C6H6(l) + (15/2)O2(g) --> 6CO2(g) + 3H2O(g)
