How do I calculate the maximum number of molecules of H2O produced when 11.5g of C4H6 (54.1g/mol) reacts with 41.8 L of O2 (32g/mol) in the following balanced equation: 2 C4H6 + 11 O2 --> 8 CO2 + 6 H2O ?

Question

Equation (balanced): 2 C4H6 + 11 O2 --> 8 CO2 + 6 H2O
Molar Mass of 11.5g of C4H6: 54.1g/mol
Molar Mass of 41.8g of O2: 32.00g/mol
Looking for: Maximum number of molecules of H2O produced

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Answer 1 · 2018-03-23T16:46:24

Let us make this a bit easier on ourselves...and rewrite the equation...we assume room temperature and atmospheric pressure...

#C_4H_6(g) + 7/2O_2(g) rarr 4CO_2(g) +3H_2O(l)#
I use the half-integral dioxygen coefficient because I find the arithmetic a bit easier this way...

#"Moles of olefin"=(11.5*g)/(54.1*g*mol^-1)=0.213*mol#.

#"Moles of dioxygen"=(PV)/(RT)=(1*atmxx41.8*L)/(0.0821*L*atm*K^-1*mol^-1xx298*K)=1.71*mol#.

And thus there is EXCESS dioxygen...and with respect to water, the equation predicts a molar quantity of...

#0.213*molxx3=0.639*mol#...i.e.

#0.639*molxx6.022xx10^23*mol^-1=3.85xx10^23# individual water molecules....