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A sample of 10.5g of Nirtogen reacts with 20.2g of Hydrogen to produce ammonia.
3H2(g)+N2(g)=2NH3(g)
a. What is the theoretical yield of ammonia?
b. What is the limiting reactant?
c. What is the excess reactant?
d. How much is the excess reactant?
e. If 4.1 g of ammonia is actually produced, what is the precent yield?​

Respuesta :

Medunno13

Answer:

Explanation:

a) The mass of the reactants is 10.5 + 20.2 = 30.7, so the theoretical yield of ammonia is also 30.7 grams (by the law of conservation of mass)

b) The gram-formula mass of diatomic nitrogen is about 28.014 g/mol, so 10.5 grams of nitrogen is 10.5/28.014 = 0.37 moles. In comparison, the gram-formula mass of diatomic hydrogen is about 2.016 g/mol, so 20.2 grams of hydrogen is about 20.2/2.016 = 10.0 moles. Dividing this by 3, we get 3.33 moles. So, nitrogen is the limiting reactant.

c) Hydrogen

d) 0.37 moles of nitrogen is consumed, so (0.37)(3) = 1.11 moles of hydrogen is consumed. This means that 3.33-1.11=2.22 moles of hydrogen are remaining, which has a mass of (2.22)(2.016)=4.47 grams.

e) (actual)/(theoretical) * 100 = (4.1)/(30.7) * 100 = 13%

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