N2 (g) + 3H2 (g) - > 2NH3 (g)

How many grams of N2 are required to produce 100.0 L of NH3 at STP

a. 6.25 g

b. 12.5 g

c. 62.5 g

d. 125.0 g

Each mole of nitrogen produces 2 moles of ammonia.

One mole of gas occues 22.4 liters under STP.

Moles of ammonia = 100/22.4

= 4.46 moles

Moles of nitrogen required = 4.64/2

= 2.32 moles

Mass = moles x Mr

Mass of nitrogen = 2.32 x 28

= 64.96 grams

I would say the answer is C.

One mole of a gas (like ammonia NH3) takes up approximately 22.41 liters at STP. Therefore, we can divide 100.0 L by 22.41 Liters/mole to get the number of moles of NH3.

100.0 L / (22.41 L/mol) = 4.462 moles NH3

Now, we know that there is one mole of N2 for every two moles of NH3, so we can get the number of moles of N2.

4.462 moles NH3 * (1 mole N2) / (2 moles NH3) = 2.231 moles N2

Now, we can multiply the number of moles of N2 by its molar mass (28.02 g/mol) to get the number of grams we need.

2.231 moles N2 * (28.02 grams N2) / (1 mole N2) = 62.52 grams N2

62.5 grams N2 are required