Fluorine gas can react with ammonia gas to produce dinitrogen tetrafluoride gas and hydrogen fluoride gas. This reaction is described by the following balanced equation:
5 F2 (g) + 2 NH3 (g) N2F4 (g) + 6 HF (g)

What mass of hydrogen fluoride gas is produced from 135 g of ammonia gas assuming that there is an ample amount of fluorine gas present for the reaction to be completed?

476g of hydrogen fluoride was produced

Explanation:

In this question, we have 2 moles of ammonia gas producing 6 moles of hydrogen Fluoride gas. This is the theoretical output we have. Let’s try and get what actually is happening!

Firstly, we calculate the number of moles of ammonia in 135g of ammonia.

Mathematically, this can be calculated by dividing the mass given by the molar mass of ammonia.

For NH3, the molar mass is 14 + 3(1) = 17g/mol where 14 and 1 are the atomic masses of nitrogen and hydrogen respectively.

The number of moles of ammonia reacted is thus 135g/17g/mol = 7.94 moles of ammonia

Theoretically, 2 moles ammonia yielded 6 moles hydrogen Fluoride, thus 7.94 moles of ammonia will yield = (7.94 * 6)/2 = 23.82 moles of Hydrogen Fluoride

we now proceed to calculate the equivalent mass of HF here.

To calculate this, we simply multiply the number of moles of hydrogen fluoride by the molar mass of hydrogen fluoride

The molar mass of hydrogen fluoride HF is 1 + 19 = 20g/mol

where 1 and 19 are the atomic masses of hydrogen and fluorine respectively.

The mass produced is thus 20 * 23.82 = 476.47 which is approximately 476g to the nearest whole number

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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.