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natajaeecarr
27.09.2019 •
Chemistry
N2 + 3h2 = 2nh3
for the equation in (2) above, you have 1000. g of n2 and 150.0 g of h2:
a. identify the limiting reactant.
b. calculate the theoretical yield in grams of nh3.
c. calculate the grams of excess reactant.
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Ответ:
The answer to your question is:
a) Hydrogen
b) 850 g of NH3
c) 300 g of N2
Explanation:
N2 + 3H2 ⇒ 2NH3
N2 = 1000 g
H2 = 150 g
a. Identify the limiting reactant.
MW N2 = 28 g
MW H2 = 2 g
MW NH3 = 17g
N2 + 3H2 ⇒ 2NH3
28g 3(2)
1000 g x
x = (1000 x 6)/ 28
x = 214.3 g of H2
Then, the limiting reactant is H2, because is necessary to have 214.3 g of H2 to react with 1000 g of N2, and there are only 150 g.
b. Calculate the theoretical yield in grams of NH3.
N2 + 3H2 ⇒ 2NH3
6g 2(17)g
150g x
x = (150 x 34) / 6
x = 850 g of NH3
c. Calculate the grams of excess reactant.
N2 + 3H2 ⇒ 2NH3
28g 6 g
x 150 g
x = (150 x 28)/ 6
x = 700 g of N2
Excess reactant = 1000g - 700g
= 300 g of N2
Ответ:
3) Rubidium reacts with rubidium nitrate to produce rubidium oxide and nitrogen gas
a- Write the chemical reaction
10Rb + 2RbNO3 → 6Rb2O + N2