# AP Chemistry

A typical bathtub can hold 94 gallons of water. Calculate the mass of natural gas that would need to be burned to heat the water for a tub of this size from 59 ◦F to 101 ◦F. Assume that the natural gas is pure methane (CH4) and that the products of combustion are carbon dioxide and water (liquid). Answer in units of g.

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1. Convert 59 F to C.
Convert 101 F to C.
Calculate q (heat) needed to raise temperature from Tinitial to Tfinal. Do that this way.
q = mass H2O x specific heat H2O x (Tfinal-Tinitial). You will need to convert 94 gallons to liters and that to mL. The number of mL will be the grams H2O assuming the density of H2O of 1.00 g/mL.

Then CH4 + 2O2 ==> CO2 + 2H2O
How much heat do you get from the rxn? That's
dHrxn = (n*dHf products)- (n*dHf reactants).

So you dHrxn for 16 g CH4. Let's call dHrxn Y.
Then 16 g x (heat needed/Y) = ? g CH4 needed.

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2. So my work then came out to be this, but proved to be wrong.

//Chemical Equation
CH4(g) + 202(g) = CO2(g) + 2H20(l)

//Gallons -> Liters
94 x 3.785 = 355.79

//Liters -> Mililiters
355.79 * 1000 = 355790

//Mililiters -> Grams (Density)
a = 355790g H2O

//Delta Temperature
b = 23 C

q = m x s x dT
q = (355790g)(4.184J/gC)(23C)
q = 34238383.28J

//Hess's law
891 kJ/mol

16.05 x 34238383.28J / 891 = 616752.022

Could you spot my mistake?

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3. I would have used 23.3 for delta T but that isn't that big a deal. Makes q about 3.47E7 J.
I have (2*H2O + CO2)-(CH4)
(2*-187.8)+(-393.5) - 74.81 and I have something like 700 kJ or so. Check that if you will but it isn't exactly 700 kJ. That's a significant difference. But a huge difference is in the last step you have J/kJ. You need to convert that 3.47E7 J to kJ. Hope this helps.

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4. I picked up the dH formation for H2O2 and not H2O. Sorry about that. So other than the 23.3 C as delta T, the other big change you need to make is the conversion of J to kJ.

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