Propane, C3H8, is a common fuel gas. Use the following to calculate the grams of propane you would need to provide 369 kJ of heat.

C3H8 (g) + 5O2 (g) --> 3CO2 (g)+ 4H2O (g) ; Delta H = -2043kJ

2,043 kJ x (grams C3H8/44) = 369 kJ

Solve for grams.

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To calculate the grams of propane needed to provide 369 kJ of heat, we can use the following steps:

1. Calculate the moles of heat provided by 369 kJ:
- 1 kJ = 1000 J, so 369 kJ = 369,000 J
- Since the chemical reaction provides -2043 kJ (exothermic), the heat provided by the reaction is -2043 kJ.
- We can use a proportion to find the moles of heat provided:
- (-2043 kJ / 1) = (x moles / 369,000 J)
- x = (-2043 kJ * 369,000 J) / 1 = -754,167 moles J

2. The balanced chemical equation shows that the molar ratio between propane (C3H8) and heat is 1:2043 (stoichiometric ratio).
- This means that for every 1 mole of propane burned, -2043 kJ of heat is released.
- Therefore, the moles of propane needed to provide 754,167 J of heat is:
- (-754,167 moles J / -2043 kJ) = 369.343 moles of propane

3. Finally, we can calculate the mass (in grams) of propane needed using its molar mass:
- The molar mass of propane (C3H8) can be calculated by adding up the atomic masses of its elements:
- C (carbon) = 12.01 g/mol * 3 = 36.03 g/mol
- H (hydrogen) = 1.01 g/mol * 8 = 8.08 g/mol
- Total molar mass of propane (C3H8) = 36.03 g/mol + 8.08 g/mol = 44.11 g/mol
- The mass of propane needed is calculated as:
- (369.343 moles * 44.11 g/mol) = 16,272.181 g (or 16,272 g to three significant figures)

Therefore, you would need approximately 16,272 grams of propane to provide 369 kJ of heat.

To calculate the grams of propane (C3H8) needed to provide 369 kJ of heat, we need to use the given balanced chemical equation and the delta H value.

The balanced chemical equation is:
C3H8 (g) + 5O2 (g) --> 3CO2 (g) + 4H2O (g)

According to the equation, the molar ratio of C3H8 to delta H is 1: -2043 kJ.

We can set up a proportion to solve for the grams of propane needed:
(369 kJ) / (-2043 kJ) = x (grams of C3H8) / (1 mole of C3H8)

First, we need to convert the given kJ value to J:
369 kJ * 1000 J/kJ = 369,000 J

Next, we need to convert the given delta H value to J:
-2043 kJ * 1000 J/kJ = -2,043,000 J

Now, we can set up the proportion and solve for x:
(369,000 J) / (-2,043,000 J) = x (grams of C3H8) / (1 mole of C3H8)

Cross-multiplying, we get:
(369,000 J) * (1 mole of C3H8) = (-2,043,000 J) * x (grams of C3H8)

Simplifying the equation:
369,000 = -2,043,000 * x

Solving for x:
x = 369,000 / -2,043,000 = -0.18

Since the number of grams cannot be negative, the answer is not physically meaningful. This suggests that the given amount of heat (369 kJ) cannot be provided by the combustion of propane (C3H8) alone.

Therefore, it is not possible to calculate the grams of propane needed to provide 369 kJ of heat based on the given information.