Anonymous
Anonymous asked in Science & MathematicsChemistry · 4 weeks ago

# need help identifying 5 errors asap?

An aqueous solution of 0.00585 mol FeCl3 is mixed with an aqueous solution of 0.700 g

NaOH to form a precipitate according to:

FeCl3(aq) + NaOH(aq)  Fe(OH)3(s) + Na +(aq) + Cl-

(aq)

a. How many grams of precipitate form if the yield of the reaction is 75.2%?

b. Assuming theoretical yield (100% yield), how many moles of Na+

(aq) that

remains in solution?

a. How many grams of precipitate form if the yield of the reaction is 75.2%?

2 amounts of 2 reactants - Need to find the limiting reactant…

Step 1:

0.00585 mol FeCl3 = 0.005850 mol Fe(OH)3

Step 2:

0.700 g NaOH x 1 𝑚𝑜𝑙𝑁𝑎𝑂𝐻

39.997 𝑔 𝑁𝑎𝑂𝐻 = 0.01750 𝑚𝑜𝑙

0.01750 mol = 0.01750 mol Fe(OH)3

Since less moles of the product are formed when using FeCl3, then Fe(OH)3 is the limiting

reactant and is completely used up.

0.00585 moles Fe(OH)3 x  106.874g Fe(OH)3/                    1mol Fe(OH)3  = 0.625 g

Actual yield = 75.2% x 0.625 g = 0.47 g

b. Assuming theoretical yield (100% yield), how many moles of Na+ (aq) that remains in

solution?

0.005850 mol FeCl3 = 0.005850 mol Na+ according to the balanced equation.

Relevance
• First mistake is that your equation is not balanced:

FeCl3(aq) + 3 NaOH(aq) --> Fe(OH)3(s) + 3 Na+(aq) + 3 Cl-

Identifying Limiting reactant:

Moles OH- = 0.700 g / 40.0 g/mol = 0.0175 mol OH-

0.00585 mol FeCl3 X (3 mol OH- / 1 mol FeCl3) = 0.01755 mol OH-

Because you do not have that much available, FeCl3 is the limiting reactant

a. 0.0175 mol OH- X (1 mol Fe(OH)3 / 3 mol OH-) X 106.9 g/mol = 0.624 g Fe(OH)3

At 75.2% yield: 0.624 X 0.752 = 0.469 g Fe(OH)3

b. Initial moles Na+ = 0.0175 mol

All of that remains in solution since it is a spectator ion in the reaction.