I have this equation Fe(H2O)6 2+ + H2C204(aq) +0.5 H20 + 2C2O4 2- (aq) + 3K+-->

Question

I have this equation
Fe(H2O)6 2+ + H2C204(aq) +0.5 H20 + 2C2O4 2- (aq) + 3K+--> H3O+ (aq) + K3[Fe(C2O4)3] 3H2O(s) + 3H2O(l)

I have 10.005g of Fe(H20)6 2+
8.3039 g oxalic acid
6.0060g of K2C2O4H20.H20
1.02 gr of H202

Calculate final mass of K3[Fe(C2O4)3] 3H2O(s)
theoretical mass of K3[Fe(C2O4)3] 3H2O(s)
percent yield K3[Fe(C2O4)3] 3H2O(s)

I know that we have to calculate the limiting reactant but im confused on how to calculate the limiting reactant for K2C2O4H20.H20 ( Professor told us to ignore the 3K+
HELPPPP LIFE SAVER if right '!!!

Explanation / Answer

Since there is no equation, I'm going to assume it's a one to one mole ratio between the two compounds: First find the molecular weight of the first compound by adding each elements atomic weight from the periodic table, I got 391.8 grams/mole but I would defiantly recheck since they are such long compounds, I could have misadded. So to find moles 11.356 g FAS/ 391.8 g/mol = .0289 moles Like I said, I'm assuming it's a 1:1 ratio so that means there is .0289 moles of PTT also. .0289 moles x 490.8 grams/moles PTT = 14.22 grams Percent yield is actual/theortical x 100% 9.376/14.22 x 100% = 66% yield.

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---