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>
Answer:
See explanation
Explanation:
The question is incomplete because the image of the alcohol is missing. However, I will try give you a general picture of the reaction known as hydroboration of alkenes.
This reaction occurs in two steps. In the first step, -BH2 and H add to the same face of the double bond (syn addition).
In the second step, alkaline hydrogen peroxide is added and the alcohol is formed.
Note that the BH2 and H adds to the two atoms of the double bond. The final product of the reaction appears as if water was added to the original alkene following an anti-Markovnikov mechanism.
Steric hindrance is known to play a major role in this reaction as good yield of the anti-Markovnikov like product is obtained with alkenes having one of the carbon atoms of the double bond significantly hindered.
Every mole of MgCl2 reacts with 2 moles of KOH, therefore the 4 moles of KOH will only react with 2 moles of MgCl2, making it the limiting reagent and therefore KOH determines how much Mg(OH)2 is produced.
