Answer:
6.75 g of HCN can be produced by the reaction
Explanation:
Complete reaction is:
2NH₃ (g) + 3O₂ (g) + 2CH₄ (g) → 2HCN (g) + 6H₂O (g)
Let's determine the moles of each reactant:
11.5 g . 1mol / 17g = 0.676 moles of ammonia
12 g . 1 mol / 32g = 0.375 moles of oxygen
10.5 g . 1mol/ 16 g = 0.656 moles of methane
Now is all about rules of three:
2 moles of ammonia reacts with 3 moles of O₂ and 2 moles of methane
0.676 moles of NH₃ may react with:
(0.676 . 3) /2 = 1.014 moles of O₂
(0.676 . 2) / 2 = 0.676 moles of methane
Both can be the limiting reactant.
3 moles of O₂ react with 2 moles of NH₃ and 2 moles of methane
0.375 moles of O₂ will react with:
(0.375 .2) / 3 = 0.375 moles
The same amount for methane, 0.375 moles
2 moles of CH₄ reacts with 3 moles of O₂ and 2 moles of NH₃
0.656 moles of methane would react with 0.656 moles of NH₃
(0.656 . 3 ) /2 = 0.437 moles of O₂ I do not have enough O₂
Oxygen is the limiting reactant → We can work with the reaction now.
Ratio is 3:2. 3 moles of oxygen produce 2 moles of cyanide
0.375 moles of O₂ may produce (0.375 .2 ) / 3 = 0.250 moles
If we convert the moles to mass → 0.250 mol . 27 g / 1mol = 6.75 g
Mol/(L s)
.
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= 
= 2.50 X