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Under appropriate conditions, nitrogen and hydrogen undergo combination reaction t0 vield ammonia:Nz (g) + 3Hz (g) ~ 2NH3If the reaction yield is 87.5%, how many mo...

Question

Under appropriate conditions, nitrogen and hydrogen undergo combination reaction t0 vield ammonia:Nz (g) + 3Hz (g) ~ 2NH3If the reaction yield is 87.5%, how many moles of Nz are needed to produce 3.00 mol of NH3?

Under appropriate conditions, nitrogen and hydrogen undergo combination reaction t0 vield ammonia: Nz (g) + 3Hz (g) ~ 2NH3 If the reaction yield is 87.5%, how many moles of Nz are needed to produce 3.00 mol of NH3?



Answers

The Haber process for making ammonia from nitrogen in the air is given by the equation $\mathrm{N}_{2}+3 \mathrm{H}_{2} \rightarrow 2 \mathrm{NH}_{3}$. Calculate the mass of hydrogen that must be supplied to make $5.00 \times 10^{2} \mathrm{kg}$ of ammonia in a system that has an $88.8 \%$ yield.

First time writing the reaction. So here I can write end to guests Plus three H 2 gas gives two. NH three. Guess now the volume of NH three B n S three can be written H seven deter H two, multiplication, two more An Edge three x 3 mol. As to which on simplification I get the volume add 4.67 l and at three. So the volume of n. S. Three that can be produced in the reaction aid 4.67 later. So this is our answer for this problem. I hope you understand how I solve this problem. I just first right the reaction after that. I just can't really deep volume off and S three in this problem.

We are given here the violence equation for the reaction between nitrogen on hydrogen to obtain ammonia. Okay, Aunt were asked to the terming the mask off ammonia being grounds that is obtained when 2.7 zero grams off hydrogen completely react with access with nitrogen in the presence off excess nitrogen. Okay, so we have the balanced equation on the two substances that I need to focus on our hydrogen and ammonia on These are the one that I'm going to use to establish be more racial on the more racial is three moles off hydrogen. We're Proview's two months, both ammonia. Okay, on this is the ratio that I'm going to use for the calculation. We have three moles off hydrogen react to produce two months of ammonia. However, we are given the mask off the amount off hydrogen in grands, then massive Kyrie in that is going to react on were asked to calculate they must Ingram's off ammonia that is going to to get obtained out off the 2.70 grams off hydrogen. Therefore, we need to convert the more racial in tow, a mask, racial, and we do that by multiplying. It's more by the molar mass. In the case of fighting, you want to apply by to times one two times one, which is the atomic weight off hydrogen. And this is grant sperm Ole. And in the case of pneumonia, we want to ply by 14 which is the atomic weight of nitrogen, plus three times one that weight of hydrogen. That's three grams promote. Then we do this leader calculation. This is six grams up here. The more will counsel, please three times to six on Dhere. It would be 17 times to 34 grants. Okay, so the mass racial is six grounds off hydrogen. We react to produce 34 grams off ammonia. This is the case when 2.7 zero grams off hydrogen react What would be the Mass Ingram's? Well, someone you. All we have to do is this little calculation 34 times 2.7. You bite your by six. That's 15 going three grounds. Both Harmonia Arab T

Using the balanced equation. If 2.70 g of hydrogen reacts. How many grams of NH three is formed, starting from 2.70 g? Reach to. We're gonna play the Moller Mouse of H two to convert two moles and then use theme or racial from the balanced chemical equation. Three moles of H two to two moles of N H three and the Mass of NH three will use the mole amounts of NH 31 more supposed to 17.31 g, and this would yield 15 2 grounds of NH three produced upon reacting to 0.70 g of hydrogen.

Multiple factors come from the coefficients of balanced chemical reactions. These coefficients and multiple factors allow us to convert between the moles of one reactant or product and the moles of another reactant or product. So if one mole end to reacts with three moles H two to produce two moles of ammonia, then if we have 1.8 moles of end to, we can calculate the moles of H two that are required. Knowing for every mole and two, we need three moles H two, so if we have 18 moles into, we need 54 moles H two. If we want to produce 0.6 moles of ammonia for every two moles of ammonia produced, we need one mole of nitrogen. So 10.6 moles of ammonia would require 0.3 moles of nitrogen. If we have 1.4 moles of hydrogen, knowing that one, sorry, the three moles of hydrogen produced two moles of ammonia, then 1.4 moles of hydrogen multiplying by the multiple factor 2 to 3 will produce 30.93 moles of ammonia.


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