Answer:
The correct answer would be : 33.8 g
Explanation:
Molar mass of ammonia,
Molar Mass = 1* Molar Mass(N) + 3* Molar Mass (H)
= 1*14.01 + 3*1.008 = 17.034 g/mol
mass(NH3)= 25.0 g (given)
number of mol of NH3,
n = mass of NH3/molar mass of NH3
=(25.0 g)/(17.034 g/mol)
= 1.468 mol
Now,
Molar mass of O2
= 32 g/mol
mass(O2)= 45.0 g
similar as ammonia
n (O2)=(45.0 g)/(32 g/mol)
= 1.406 mol
Balanced chemical equation is:
4 NH3 + 5 O2 ---> 4 NO + 6 H2O
1.83456 mol of O2 is required for 1.46765 mol of NH3
by the calculation we have only 1.40625 mol of O2
Thus, the limiting agent will be - O2
now the Molar mass of NO,
= 1*14.01 + 1*16.0
= 30.01 g/mol (similar formula used for NH3)
Balanced equation :
mol of NO formed = (4/5)* moles of O2
= (4/5)×1.40625 (from above calculation)
= 1.125 mol
mass of NO = number of moles × molar mass
= 1.125*30.01
= 33.8 g
Thus, the correct answer would be : 33.8 g
The amount of nitrogen oxide that can be formed in the given mass is 44.12 g.
The given parameters;
The reaction of the ammonia and oxygen is written as follows;
[tex]4NH_3(g) \ + \ 5O_2 (g) \ --> \ 4NO (g) \ + \ 6H_2O(g)\\\\[/tex]
Molar mass of NH₃ = (14) + (3 x 1) = 17 g/mol
Molar mass of NO = (14) + 16 = 30 g/mol
4(17 g/mol) of NH₃ ------------------ 4(30)
25 g/mol of NH₃ --------------------- ?
[tex]= \frac{4(30) \times 25}{4(17)} \\\\= 44.12 \ g[/tex]
Thus, the amount of nitrogen oxide that can be formed in the given mass is 44.12 g.
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