Can anyone solve this chemistry problem?

Question

C6H12O6(aq) + 6 O2(g) ------- 6 CO2(g) + 6 H2O(l)

Calculate the volume of CO2 produced at a body temperature(37Cdegrees) and 738 mmHg when 24.5-g of glucose is consumed in this reaction?

Answer 1

Its clear that you know the theory.
So do your own home work yourself.

Answer 2

I'm not going to do the whole problem for you but I will tell you how to do it....

This is a problem using stoichiometry and the Ideal Gas Law
PV = nRT
1 - Convert temperature into Kelvin (37+273)
2 - Convert mmHg to atm's (738/760)
convert 24.5-g to moles (24.5g / (6+ 32 g/mol)
now that you have the moles, there are 6 mols of CO2 in the equation, so multiply this by six
3- now that you have your number of moles (n), your pressure (p), your temperature (t), you can use the equation PV = nRT

V = nRT / P (R is a constant, 8.12)

Answer 3

This is a limiting reactant problem
you have 24.5 grams of glucose and it takes one mole of glucose to form 6 moles of CO2.

glucose has 180 g/mol
24g/180g/mol = .136111moles

CO2 has 44g/mol

0.136111mol X 6 mol CO2
----------------------------------
..................... 1 mole glucose

= 0.8167 mol CO2

V=nRT/P is the universal gas law since CO2 is a gas
where n= # of moles
where R is 0.8206
you need to change temperature to Kelvin

276 + 37= 313K
and 738 mmHg to atm
738/760
=0.971atm

=.816 X .08206 X 310K
--------------------------------
.......................971atm

V=21.38L

Answer 4

MW glucose = 180 g/mol (not 160)

24.5 g C6H12O6 / 180 g/mol = 0.136 mol

0.136 mol C6H12O6 * 6 moles CO2 / 1 mole C6H12O6 = 0.816 moles CO2

V = nRT/P = 0.816 moles * 0.08206 * 310 K / 0.971 atm = 21.38 L