Question

Please help! How many grams of N2(g) can be made from 9.05(g) of NH3 reacting with 45.2(g) of CuO?

Answer 1

2 NH3 + 3 CuO = 3 Cu + 3 H2O + N2

Answer 2

my teacher also gave this along with the problem 2NH3(g)+3CuO(s)---N2(g)+3Cu()+3H2O(l)

Answer 3

Please help! How many grams of N2(g) can be made from 9.05(g) of NH3 reacting with 45.2(g) of CuO? 2NH3(g)+3CuO(s)---N2(g)+3Cu()+3H2O(l)

Answer 4

my teacher only want the number I calculated

Answer 5

Ok, so let's find the number of moles of NH3 and CuO first. Do you know how to do that?

Answer 6

please explain.

Answer 7

ok would NH3 be 17.03 g/mol

Answer 8

and CuO would be 79.55 g/mol

Answer 9

n(NH3) = 9.05/17.03 = 0.531415 mol
n(CuO) = 45.2/79.55 = 0.568196 mol
Since 3 CuO = 2NH3, CuO is the limiting reagent.
Effective n(NH3) = 0.568196*2/3 = 0.378797 mol
Since N2 = 2NH3
n(N2) = 0.378797/2 = 0.1893987 mol
m(N2) = 0.1893987*28.0 ≈ 5.30 g

Answer 10

where did you get the 9.05 from?