C4H10(g) + 13/2 O2> 4CO2(g)+ 5H2O(g) Hrxn -2658Kj

What mass of butane in gms is necessary to produce 1.5 times ten ^3Kj of heat? What mass of CO2 is produced?

To answer these questions, we need to use the given balanced chemical equation and the associated enthalpy change (Hrxn).

1. Calculating the mass of butane:
- The balanced chemical equation tells us that 1 mole of butane (C4H10) produces -2658 Kj of heat.
- We want to produce 1.5 x 10^3 Kj of heat. So we can set up a ratio:

-2658 Kj x g butane 1.5 x 10^3 Kj
----------- = ---------------------
1 mole 1 g ?

- To solve for g butane, we can set up a cross multiplication equation:

(-2658 Kj) x (1 g) = (1.5 x 10^3 Kj) x (x g butane)

- Now solve for x:

x g butane = ((-2658 Kj) x (1 g)) / (1.5 x 10^3 Kj)
x g butane = -1.772 g

Therefore, you would need approximately 1.772 grams of butane to produce 1.5 x 10^3 Kj of heat.

2. Calculating the mass of CO2 produced:
- From the balanced chemical equation, we see that 1 mole of butane (C4H10) produces 4 moles of carbon dioxide (CO2).
- Since we now have the mass of butane in grams, we can convert it to moles using the molar mass of butane (58.12 g/mol).
- Set up the mole-mole ratio between butane and CO2:

(1.772 g butane) x (1 mol butane / 58.12 g butane) x (4 mol CO2 / 1 mol butane) x (44.01 g CO2 / 1 mol CO2)

- Calculate this expression:

(1.772 x 4 x 44.01) / 58.12 g CO2 = 5.362 g CO2

Therefore, approximately 5.362 grams of CO2 would be produced.