What is the molar mass of Fe(NH_4)_2(SO_4)_2 6H_2O(s) Using the molar mass of Fe

Question

What is the molar mass of Fe(NH_4)_2(SO_4)_2 6H_2O(s) Using the molar mass of Fe(NH_4)_2(SO_4)_2 middot 6H_2O(s) and the amount used in your experiment. 1 calculate the number of moles of Fe(NH_4)_2(SO_4)_2-6H_2O(s) you used in this experiment. According to the balanced chemical reaction, for each mole of Fe(NH_4)_2(SO_4)_2 - 6H_2O(s) consumed in the reaction, 1 mole of K_3[Fe(C_2O_4)_3]- 3H_2O is produced. Based on your result to analysis question 2, how many moles of should be produced in the experiment? The molar mass of is 491.1 g/mol. Convert the theoretical moles produced found in question 3 into theoretical grams produced.

Explanation / Answer

1)

Fe(NH4)2(SO2)2*6(H2O)

Fe = 1x

NH4 = 2x

SO4 = 2x

H2O = 6x

then

Fe = 1x = 1*55.5 = 55.5

NH4 = 2x = 2*18 = 36

SO4 = 2x = 2*96 = 192

H2O = 6x = 6*18 = 108

Total mass = 55.5+36+192+108 = 391.5

Molar MAss = 391.5 g/mol

2)

if m = 5.13 g of Fe(NH4)2(SO2)2*6(H2O)

then

mol = mass/MW = 5.13 /391.5 = 0.0131034 mol of Fe(NH4)2(SO2)2*6(H2O)

3)

Since ratio is 1:1 then

0.0131034 mol of Fe(NH4)2(SO2)2*6(H2O) produces 0.0131034 mol of K3[Fe(C2O4)3]*3H2O

4)

MW = 491.1 g/mol

so

m = mol*MW = 491.1*0.0131034 = 6.43507974 g of K3[Fe(C2O4)3]*3H2O

NOTE: consider posting all other questions in another set of Q&A. We are not allowed to answer to multiple questions in a single set of Q&A

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.