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Question

50.0 mL of 0.200 M solution of aluminum nitrate was mixed with 200 mL of 0.180 M solution of sodium carbonate.

 

Balance the equation, give the states of the products, and determine the Limiting Reactant. 

Answer 1

2Al(NO₃)_{3 (ag)} + 3Na₂CO_{3 (aq)} --->Al₂(CO₃)_{3 (s)} + 6Na(NO₃₎ (aq)

0.01 moles Al(NO₃)₃ 0.036 moles Na₂CO₃. From the balanced equation, we get molar ratios. 2 moles aluminum nitrate make 1 mole of aluminum carbonate whereas 3 moles of sodium carbonate make 1 mole of aluminum carbonate. So 0.01 m,oles of sodium carbonate yield 0.005 moles of aluminum carbonate. 0.036 moles of sodium carbonate yield 0.012 moles of aluminum carbonate. We make less product from the aluminum nitrate; it will run out first. it is the limiting reagent. The number of moles of each reactant is calculated by multiplying the molarity by the volume. it is helpful to convert mL to liters.

Answer 2

50.0 mL of 0.200 M solution of aluminum nitrate was mixed with 200 mL of 0.180 M solution of sodium carbonate.

moles=M*V

=0.200mol/L *0.050L=0.01 moles Aluminum nitrate

moles =M*V

=0.180mol/L*0.200L =0.036 moles Sodium carbonate

The balanced chemical equation is

2Al(NO3)3(aq) +3Na2CO3(aq) ==>Al2(CO3)3(s) +6NaNO3(aq)

mole ratio of Al(NO3)3 :Al2(CO3)3 is 2:1

0.01 moles Aluminum nitrate produces 0.005 moles Al2(CO3)3. (Least)

mole ratio is Na2CO3:Al2(CO3)3 is 3:1

0.036 moles Sodium carbonate produces 0.012 mol Al2(CO3)3 (more)

Least quantity is produced by the Limiting reactant.

So, the limiting reactant is Al(NO3)3.