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25 ml of 2.4 M K2(CO3) is mixed with 35 ml of 2.0 M Al(NO3)3. (a) Evaluate the moles for each reactant (b) indicate the limiting reactant (c) how many moles of Al2(CO3)3 are formed?

Sagot :

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Answer:

a) 0.06 mol K₂(CO₃) and 0.07 mol Al(NO₃)₃

b) The limiting reactant is K₂CO₃

c) 0.03 moles of Al₂(CO₃)₃

Explanation:

The balanced reaction between K₂CO₃ and Al(NO₃)₃ is the following:

3 K₂CO₃ + 2 Al(NO₃)₃ → Al₂(CO₃)₃ + 6 KNO₃

(a) We can calculate the moles for each reactant as the product of the molarity (M) and the volume (V) in liters.

K₂(CO₃): M = 2.4 mol/L , V = 25 mL x 1 L/1000 mL = 0.025 L

⇒ moles K₂(CO₃) = M x V = 2.4 mol/L x 0.025 L = 0.06 mol K₂(CO₃)

Al(NO₃)₃: M = 2.0 mol/L , V = 35 mL x 1 L/1000 mL = 0.035 L

⇒ moles Al(NO₃)₃ = M x V = 2.0 mol/L x 0.035 L = 0.07 mol Al(NO₃)₃

(b) According to the balanced equation, the stoichiometric ratio is 3 moles K₂CO₃/2 moles Al(NO₃)₃. We have 0.07 moles of Al(NO₃)₃, so we multiply the moles of Al(NO₃)₃ by the stiochiometric ratio to calculate how many moles of K₂CO₃ are needed:

0.07 mol Al(NO₃)₃ x 3 mol K₂CO₃/2 mol Al(NO₃)₃ = 0.105 moles of K₂CO₃

We need 0.105 moles of K₂CO₃ but we have only 0.06 moles of K₂CO₃. Therefore, the limiting reactant is K₂CO₃.

(c) We use the limiting reactant to calculate how many moles of product (Al₂(CO₃)₃) are formed. According to the equation, 2 moles of K₂CO₃ produce 1 mol of Al₂(CO₃)₃, thus the stoichiometric ratio is 1 mol Al₂(CO₃)₃/2 moles K₂CO₃. We have 0.06 moles of K₂CO₃, so the number of moles of Al₂(CO₃)₃ will be:

0.06 moles K₂CO₃ x 1 mol Al₂(CO₃)₃/2 moles K₂CO₃ = 0.03 moles of Al₂(CO₃)₃

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