Question

11. In a reaction from number 10, 65.0g of Ni(NO3)2 is reacted with 58.0g KOH. Which is

the limiting reactant? Show your work for credit. (4pts)

Answer 1

Ni(NO3)2 is the limiting reactant.

Step-by-step explanation:

- First, we balance the equation...

Ni(NO3)2 + 2 KOH ---> 2 KNO3 + Ni(OH)2

- Second, we find the moles of each substance...

65g Ni(NO3)2 / 182.703g Ni(NO3)2 = 0.356 mol Ni(NO3)2

58g KOH / 56.1056g KOH = 1.034 mol KOH

- Third, to make the molar ratio equal to each other for comparison, we either multiply KOH by 1/2 or multiply Ni(NO3)2 by 2 to compare the number of moles; because the Ni(NO3)2 to KOH molar ratio is 1 to 2. Note that the multiplication of moles is only for comparison. We do not use these multiplied values. We use the values from step 2...

0.356 mol Ni(NO3)2 * 2 = 0.712 mol Ni(NO3)2

0.712 mol Ni(NO3)2 < 1.034 mol KOH ... Ni(NO3)2 is the limiting reactant.