Answer:
11.6g of NH₃(g) have to react
Explanation:
For the reaction:
4 NH₃(g) + 5 O₂(g) → 4 NO(g) + 6 H₂O(g) ΔH = -905kJ
<em>4 moles of ammonia produce 905kJ</em>
Thus, if you want to produce 154kJ of energy you need:
154kJ × (4 mol NH₃ / 905kJ) = <em>0.681moles of NH₃. </em>In mass -Molar mass ammonia is 17.031g/mol-
0.681mol NH₃ × (17.031g / mol) = <em>11.6g of NH₃(g) have to react</em>
C:H:O = 40/12 : 6,7/1 : 53,3/16 = 3,33 : 6,7 : 3,33 ≈ 1 : 2 : 1
CH₂O
because it has ns1 electron configuration like the alkali metals