How many grams of nh3 can be produced from 3.80 mol of n2 and excess h2?

1 answer:

4 0

The balanced equation :N2 + 3H2 --> 2NH3

1) Let's find the molar mass of NH3 because it will be needed later.
To determine the molar mass you have to consult the periodic table and see the atomic mass of each element. If the element is repeted, for example 3 times, you multiple the atomic mass per 3.
NH3= 14 + 1 x 3= 17 g/mol

2) By looking at the balanced equation, you can see that 2 mol of NH₃ were produced with only one mol of N2. How much NH3 will result when we have 3,80 mol of N2?
So now we solve it with a porportion:

2 mol NH3-------1 mol of N2
x --------3.80 of N2

x= 2 x3,8 = 7,6mol NH3

3) the only thing left to do is convert the moles obtained, to grams. We can use the formula:
nº of moles x molar mass= mass
7,6mol NH3 x 17 g/mol= mass of NH3
mass= 129,2 grams of NH3

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Read 2 more answers

When 5.0 mol Al react with 13 mol HCl, what is the limiting reactant, and how many moles of AlCl3 can be formed? (2 points) 2Al

barxatty [35]

Well, we need to find the ratio of Al to the other reactant.

Al:HCl = 1:3

--> this means that for every 1 Al used, you have to use 3 HCl.

5*3 = 15 moles of HCl needed to fully react with 5 moles of Al. Since 13<15, HCL is the limiting reactant.

The ratio of HCl:AlCl = 3:1

13/3 = 4.3333...

The final answer is HCl is the limiting reactant with 4.3 moles of AlCl3 able to be produced so choice B.