heat problems?

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CH4 + 3Cl2 ----> CHCl3 + 3HCl
^H degrees for the reaction is -336KJ. How do you calculate the heat when 23.0g of HCl are formed?

2007-07-25 18:20:33 · 2 answers · asked by Matthew H 2 in Science & Mathematics ➔ Chemistry

2 answers

I presume you mean -336kj/mol.

first determine moles of HCl produced:

23.0g/[36.45g/mol] = 0.63 mole.

But for each mole of HCl formed you only make 1/3 mole of CHCl3; so you need to divide by 3:

0.63mole HCl/3 = 0.21mol CHCl3

0.21mol CHCl3 formed x -336kj/mol = -70.6kj

2007-07-25 18:42:48 · answer #1 · answered by Flying Dragon 7 · 0⤊ 0⤋

Can you dumb that down a bit for us normies?

2007-07-26 01:27:56 · answer #2 · answered by Anonymous · 0⤊ 0⤋