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this is limiting reactant
Question 3 of 10
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23.227 grams of NO is formed upon the complete reaction of 31.0 grams of oxygen gas with excess ammonia.
Explanation:
The balanced chemical equation is:
4NH3 + 5O2 ⇒ 4NO + 6H20
Number of moles will be calculated.
31 grams of oxygen is given
atomic weight of oxygen gas i.e O2 = 32gm/mole
The molar mass of oxygen gas is 32 grams/mole which is equal to one mole of the molecule.
Number of moles (n) =
n=
n = 0.968 moles of oxygen are given for the reaction to occur.
From the balanced chemical equation it can be seen
5 moles of oxygen yielded 4 moles of NO
So, 0.968 moles of oxygen yield x moles of NO
=
4 × 0.968 =5x
x = 0.774 moles of NO will be formed.
To calculate the mass of NO
mass= atomic mass x number of moles
atomic mass of NO is 30.01 grams/mole
putting the values in the formula:
mass = 30.01 x 0.774
= 23.227 grams of NO is formed.