What mass of ammonia is left in excess?
5
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g
The production capacity for acrylonitrile (C3H3N) in the United States exceeds 2 million pounds per year. Acrylonitrile, the building block for polyacrylonitrile fibers and a variety of plastics, is produced from gaseous propylene, ammonia, and oxygen.
2 C3H6(g) + 2 NH3(g) + 3 O2(g) → 2 C3H3N(g) + 6 H2O(g)
What mass of oxygen is left in excess?
? g
What mass of ammonia is left in excess?
? g
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To determine the mass of ammonia that is left in excess, we need to compare the stoichiometric ratio of ammonia to the limiting reactant in the balanced chemical equation.
From the balanced equation:
2 C3H6(g) + 2 NH3(g) + 3 O2(g) → 2 C3H3N(g) + 6 H2O(g)
We can see that the stoichiometric ratio between ammonia (NH3) and acrylonitrile (C3H3N) is 2:2, which means that 2 moles of NH3 are required to produce 2 moles of C3H3N.
Let's assume that the amount of propylene (C3H6) and oxygen (O2) is in excess. This means that after the reaction, all the propylene and oxygen have been used up, leaving some ammonia in excess.
Now, we have to consider the molar ratio of ammonia (NH3) to acrylonitrile (C3H3N) to determine the amount of ammonia in excess.
As we know, the molar mass of ammonia (NH3) is 17 g/mol, and the molar mass of acrylonitrile (C3H3N) is 53 g/mol.
1 mole of NH3 = 17 g
1 mole of C3H3N = 53 g
Since the stoichiometric ratio is 2:2, we can write a proportion:
2 moles of NH3 = 2 moles of C3H3N
X moles of NH3 = 53 g of C3H3N
Now, we can solve for X (moles of NH3):
X = (53 g of C3H3N) * (2 moles of NH3 / 2 moles of C3H3N)
X = 53 g of C3H3N
So, the mass of ammonia left in excess is equal to the molar mass of acrylonitrile, which is 53 g.