How many kilocalories are involved in the production of 33.68 g of nh3?

I believe that the balanced chemical equation is:

3H2(g)+N2(g)→2NH3(g), ΔH = -21.9 kcal

We can see that 21.9 kcal of heat is released for every 2 moles of NH3.

The molar mass of NH3 is 17 g/mol, hence:

moles NH3 = 33.68 g / (17 g/mol)

moles NH3 = 1.98 mol

The total energy released is:

total heat = (-21.9 kcal / 2 mol) * 1.98 mol

total heat = -21.70 kcal

in scientific notation, the value of the  speed of light in a vacuum

  will be written as 2.995 * 10^8. when this value is represented in the form of

  a * 10^b, then, a = 2.995 and b = 8.

the value of the speed of light in a vacuum given above can also be rounded up to approximately 300000000. in this case, the scientific notation will be written as 3 *10^8.

the answer is 303.15 or 303