Solid sodium reacts with an aqueous solution of aluminum nitrate according to th

ID: 713667 • Letter: S

Question

Solid sodium reacts with an aqueous solution of aluminum nitrate according to the following balanced equation:

3 Na (s) + Al(NO3)3 (aq) à 3 NaNO3 (aq) + Al (s)

Complete the following series of question to determine how much solid aluminum will form when 2.3 grams of Na is added to 200 mL of 0.50 M Al(NO3)3.

a: Determine the number of moles of Na (s) in 2.3 grams of Na. ANS: 0.100 mol

b: Determine the number of moles of Al(NO3)3 in 250 mL of 1.08 M Al(NO3)3. ANS: 0.27 mol

c: Determine the limiting reactant when the above quantities are mixed. Be sure to show your work and explain your reasoning. ANS: Na limits

d: Calculate the theoretical yield of Al when the above reactant quantities are mixed. ANS: 0.899 g Al

3 Na (s) + Al(NO3)3 (aq)-----------------> 3 NaNO3 (aq) + Al (s)

a. no of moles of Na = W/G.A.Wt

= 2.3/23 = 0.1 moles

b. no of moles of Al(NO3)3 = molarity * volume in L

= 1.08*0.25 = 0.27 moles

c. 3 Na (s) + Al(NO3)3 (aq)-----------------> 3 NaNO3 (aq) + Al (s)

1 moles of Al(NO3)3 react with 3 moles of Na

0.27 moles of Al(NO3)3 react with = 3*0.27/1 = 0.81 moles of Na

Na is limiting reactant

d. 3 moles of Na react with Al(NO3)3 to gives 1 mmoles of Al

0.1 moles of Na react with Al(NO3)3 to gives = 0.1*1/3 = 0.033 moles of Al

mass of Al = no of moles * gram atomic mass

= 0.033*27 = 0.899g of Al

theoretical yield of Al = 0.899g

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