Solution A contains 5.00 g of AI(C_2H_3O_2)_3 and solution B contains 5.00 g of
ID: 1062889 • Letter: S
Question
Solution A contains 5.00 g of AI(C_2H_3O_2)_3 and solution B contains 5.00 g of Na_2CrO_4. Both solutions are clear and colorless, but when you mix them together a precipitate forms. You collect the precipitate on a filter paper and measure its mass to be 3.25 g. Write the balanced molecular and net ionic equations for this reaction. Determine the limiting reactant and the mass of the precipitate that could be generated. Calculate the percent yield. A solid powder sample is a mixture of solid NaCI and solid Na_2CO_3, but the exact percentage of each in the powder is not known. To determine the percentage of these Na_2CO_3, you decide to react some of the dissolved sample with a standardized solution of HCI: Na_2CO_(aq) + 2 HCl_(aq) rightarrow 2 NaCl(aq) + H_2O_(1) + CO_2(aq) Calculate the volume of concentrated HCI (12.0 M) needed to produce 500.0 mL of 0.10 M HCI. To more accurately determine the concentration of the diluted HCI solution, you react it with 0.1056 g of 100.0% pure Na_2CO_3 dissolved in 50 mL of deionized water; calculate the concentration of the HCI solution, if it takes 19.27 mL to completely react with the pure Na_2CO_3. You dissolve 0.2655 g of the sample powder in 100 mL of deionized water; 22.15 mL of the Hcl solution is needed to completely react the Na_2CO_3 in the sample. Calculate the mass percentage of Na_2CO_3 in the sample. Balance the following redox reaction: Hg_2^2+ + Fe^3+ Fe^2+ + Hg^2+Explanation / Answer
1a) Write down the reaction first:
Al(C2H3O2)3 + Na2CrO4 ------> Al2(CrO4)3 + NaC2H3O2
Start by balancing the Al atoms on both sides:
2 Al(C2H3O2)3 + Na2CrO4 ------> Al2(CrO4)3 + NaC2H3O2
Next balance the CrO4 radicals on both sides:
2 Al(C2H3O2)3 + 3 Na2CrO4 ------> Al2(CrO4)3 + NaC2H3O2
Finally balance Na atoms and C2H3O2 radicals on both sides and mention the states. The balanced chemical equation is
2 Al(C2H3O2)3 (aq) + 3 Na2CrO4 (aq) ------> Al2(CrO4)3 (s) + 6 NaC2H3O2 (aq) (balanced)
Decompose the compounds into ions and write
2 Al3+ (aq) + 6 C2H3O2- (aq) + 6 Na+ (aq) + 3 CrO42- (aq) -----> Al2(CrO4)3 (s) + 6 Na+ (aq) + 6 C2H3O2- (aq)
Cancel out common ions from both sides to get the net ionic equation as
2 Al3+ (aq) + 3 CrO42- (aq) ------> Al2(CrO4)3 (s)
b) Molar mass of Al2(C2H3O2)3 = 204.114 g/mol and molar mass of Na2CrO4 = 161.97 g/mol.
TO determine the limiting reactant, find out which reactant produces the lower yield of the product.
Molar mass of Al2(CrO4)3 = 401.94 g/mol.
Al(C2H3O2)3: [5.00 g Al(C2H3O2)3]*[1 mole Al(C2H3O2)3/204.114 g Al(C2H3O2)3]*[1 mole Al2(CrO4)3/2 mole Al(C2H3O2)]*[401.94 g Al2(CrO4)3/1 mole Al2(CrO4)3] = 4.9229 g 4.923 g Al2(CrO4)3
Na2CrO4: [5.00 g Na2CrO4]*[1 mole Na2CrO4/161.97 g Na2CrO4]*[1 mole Al2(CrO4)3/3 mole Na2CrO4]*[401.94 g Al2(CrO4)3/1 mole Al2(CrO4)3] = 4.1359 g 4.136 g Al2(CrO4)3
Since Na2CrO4 produces a lower yield of the product, hence Na2CrO4 is the limiting reactant and the yield of the products, Al2(CrO4)3 is 4.136 g (ans).
c) Percent yield = (Actual yield/Theoretical yield)*100 = (3.25 g/4.136 g)*100 = 78.578 78.58% (ans).
2a) Use the dilution equation as
V1*S1 = V2*S2 where V1 = unknown, S1 = 12.0 M, V2 = 500 mL and S2 = 0.10 M. Therefore,
V1*(12.0 M) = (500 mL)*(0.10 M)
====> V1 = (500 mL)*(0.10)/(12.0) = 4.1666 mL 4.167 mL (ans).
b) Look at the balanced equation:
Na2CO3 (aq) + 2 HCl (aq) -----> 2 NaCl (aq) + H2O (l) + CO2 (g)
Moles Na2CO3: Moles HCl = 1:2 as per the balanced stoichiometric equation.
Molar mass of Na2CO3 = 105.988 g/mol.
Therefore, moles of Na2CO3 = (0.1056 g Na2CO3)*(1 mole/105.988 g Na2CO3) = 9.963*10-4 mole.
As per the balanced chemical equation, moles of HCl required for neutralizing Na2CO3 = (9.963*10-4 mole Na2CO3)*(2 mole HCl/1 mole Na2CO3) = 1.9926*10-3 mole HCl. The volume of HCl taken = 19.27 mL.
Therefore, concentration of HCl = moles of HCl/volume of HCl in L = (1.9926*10-3 mole)/[(19.27 mL)*(1 L/1000 mL)] = (1.9926*10-3 mole)*1000/(19.27).(1 L) = 0.1034 mol/L 0.103 M (ans).
c) Concentration of HCl = 0.103 mol/L and the volume of HCl required for neutralizing Na2CO3 in the sample = 22.15 mL.
Therefore, moles of HCl required for the neutralization = (22.15 mL)*(1 L/1000 mL)*(0.103 mol/L) = 2.28145*10-3 mol.
Moles of Na2CO3 neutralized = (2.28145*10-3 mole HCl)*(1 mole Na2CO3/2 mole HCl) = 1.140725*10-3 mole.
Since molar mass of Na2CO3 = 105.988 g/mol, therefore, mass of Na2CO3 in the sample = (1.140725*10-3 mole)*(105.988 g/1 mol) = 0.1209 g.
Therefore, the given sample had 0.1209 g Na2CO3.
Percentage of Na2CO3 in the sample = (mass of Na2CO3/mass of sample)*100 = (0.1209 g/0.2655 g)*100 = 45.5367% 45.54% (ans).
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