Byron S. answered • 10/09/14

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You're given the concentrations and volumes of the two solutions.

(concentration) = (moles)/(volume)

So for a), you have 0.01M CoCl

_{3}= n / (0.010 L) and can solve for nFor b), you have 0.06M C

_{2}H_{8}N_{2}= n / (0.010 L) and again can solve for n1CoCl

_{3}+ 3NH_{2}CH_{2}CH_{2}NH_{2}→ 2HCl +*1*Co(NH_{2}CH_{2}CH_{2}NH_{2})_{3}Cl_{3}The chemical equation you're given states that you need the ratio of 1 mol CoCl

_{3}to 3 mol C_{2}H_{8}N_{2}to make the products. Check your answers for a) and b). Triple a) and compare that to b). If 3*a) is more than b), then b) runs out first, and it's the limiting reactant. If 3*a) is less than b), then a) runs out first, and is the limiting reactant.Take your limiting reactant, and use the equation to determine how much of the product is formed. CoCl3 forms the product at a 1:

*1*rate. C2H8N2 forms the product at a 3:*1*rate.
Byron S.

10/09/14