Question

A student carried out the synthesis described in this experiment. In the first step of the synthesis, the student combined 4.987 g of Fe(NH4)2(SO4)2·6H2O with 18.14 mL of 1.1 M H2C2O4. In the second step of the synthesis, the student added 20.5 mL of saturated K2C2O4, 21.2 mL of 3% H2O2, and 9.9 mL of 1.1 M H2C2O4 to the solid FeC2O4·2H2O produced in the first step. If H2C2O4 were the limiting reagent in the first step of the reaction, how many moles of FeC2O4·2H2O(s) would be created in the first step?

Answer 1

`n_( FeC_2O_4\ 2H_2O)=0.020molFeC_2O_4\ 2H_2O`

Step-by-step explanation:

Hello.

In this case, since the chemical reaction for the first step is:

`Fe(NH_4)_2(SO_4)_2\ 6H_2O + H_2C_2O_4 \rightarrow FeC_2O_4\ 2H_2O + H_2SO_4 + (NH_4)_2SO_4 + 4H_2O`

Whereas we can see a 1:1 molar ratio between FeC2O4·2H2O(s) and H2C2O4, thus, we compute the moles of yielded FeC2O4·2H2O(s) in the first step as shown below:

`n_( FeC_2O_4\ 2H_2O)=0.01814L*1.1(molH_2C_2O_4 )/(L) *(1molFeC_2O_4\ 2H_2O)/(1molH_2C_2O_4) \\\\n_( FeC_2O_4\ 2H_2O)=0.020molFeC_2O_4\ 2H_2O`

Best regards.