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.
Answer:
6.43 moles of NF₃.
Explanation:
The balanced equation for the reaction is given below:
N₂ + 3F₂ —> 2NF₃
From the balanced equation above,
3 moles of F₂ reacted to produce 2 moles of NF₃.
Finally, we shall determine the number of mole of nitrogen trifluoride (NF₃) produced by the reaction of 9.65 moles of Fluorine gas (F₂). This can be obtained as follow:
From the balanced equation above,
3 moles of F₂ reacted to produce 2 moles of NF₃.
Therefore, 9.65 moles of F₂ will react to to produce = (9.65 × 2)/3 = 6.43 moles of NF₃.
Thus, 6.43 moles of NF₃ were obtained from the reaction.
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