SiH4(g) + 2 F2(g) SiF4(g) + 4 HF(g)

If 18.9 kJ are released when 1.10 g of F2 reacts with an excess of SiH4, what is the heat of the reaction when 1.20 moles of F2 react?

To find the heat of the reaction when 1.20 moles of F2 react, we need to use the concept of stoichiometry and the given information about the heat released when 1.10 g of F2 reacts.

Given:
H2 + F2 -> 2 HF (1 mole of F2 releases 572 kJ of heat)

1. Convert the given mass of F2 to moles:
Using the molar mass of F2 (fluorine gas: 2 x 19.0 g/mol = 38.0 g/mol), we can determine the number of moles:
1.10 g F2 x (1 mol F2 / 38.0 g F2) = 0.029 mol F2

2. Determine the heat released for 0.029 moles of F2:
Since 1 mole of F2 releases 572 kJ of heat, we can set up a proportion:
1 mole F2 releases 572 kJ
0.029 moles F2 releases x kJ

Cross multiplying gives us:
1 x x = 572 x 0.029
x = 16.6 kJ

Therefore, when 1.10 g of F2 reacts, 18.9 kJ are released, and when 0.029 moles of F2 reacts, 16.6 kJ are released.

3. Calculate the heat of the reaction when 1.20 moles of F2 react:
Using a proportion, we can set up the following equation:
0.029 moles F2 releases 16.6 kJ
1.20 moles F2 releases x kJ

Cross multiplying gives us:
0.029 x = 1.20 x 16.6
x = 608 kJ

Therefore, when 1.20 moles of F2 react, 608 kJ of heat are released.

The only way to compare 1.10g to 1.20 moles is to change one of them to grams/moles.

1.20moles F2=1.2*38=45.6 grams.

Heat= 18.9kJ * 45.6/1.10