How many grams of H2 are needed to produce 10.06 g of NH3?

Answer 1

The formula reaction for NH3 when using N2 and H2 is:

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

Now, first step is to find the moles of the H2 reactant. This is found via (grams of reactant)/(molar mass of reactant).

There are 10 grams, and the molar mass of H2 is approximately 2.016. Therefore, the equation should look like:

10/2.016.

This yields a value of ~4.9606 moles of H2.

Now, you use the molar ratio from the reactant to the product to determine how many moles of product were yielded.

According to the reaction, three moles of H2 are required to produces 2 moles of NH3. So, the mole ratio is 2/3.

Multiply the number of moles of H2 with the molar ratio to determine the moles of NH3.

4.0606 * 2/3 = 3.3071 moles of NH3.

Multiply the number of moles with the molar mass of NH3 (17.0306), and voila!

3.3071 * 17.0306 = 56.3216 grams.

Now, if your teacher is feeling like a stickler about significant figures, than that value should be rounded to 56 grams of NH3.

Answer 2

Ammonia is formed by the following:

N2 + 3H2 --> 2NH3

(1) Convert grams of ammonia to moles of ammonia using its molecular weight (17 g /mole)

10.14 g NH3 / (17 g NH3/mole NH3) = 0.597 moles NH3

(2) Now, using the reaction from above we see that for every mole of ammonia formed that 3 moles of hydrogen gas are used. Assuming that the reaction is in stoichemtric proportions you can figure out how many moles of hydrogen were needed to produce 0.597 moles of NH3:

0.597 moles NH3 * (3 moles H2 / 1 mole NH3) = 1.789 moles H2

(3) Now using the molecular weight of hydrogen gas (H2) , which 2 grams/mole, multiply this by the number of moles of H2 to get your answer:

1.789 moles H2 * (2 grams H2 / 1 mole H2) = 3.58 grams H2

Answer 3

The amount of grams needed of H2 are 0.597 to produce 10.30 grams of NH3.

Answer 4

There are 0.558 NH3 moolesin that weight. You need 1.6766 H2 grams.

Answer 5

The answer is 1,77g hydrogen.