Question

An iron ore sample weighing 0.5562 g is dissolved HCl (aq), and the iron is obtained as Fe2+ in solution. This solution is then titrated with 28.72 mL of 0.04021 M K2Cr2O7 (aq). What is the percent by mass iron in the ore sample? How do I get to the equation

6Fe^2+ + 14H^+ + Cr2O7^2- ----> 6Fe^3+ + 2Cr^3+ + 7H2O ??? I am just generally confused on how to start this and how to do this step by step. Would really appreciate a very detailed, and step by step answer. I've found similar questions, but I got stuck on the very 1st step, and that's writing out an equation.

I need this answered ASAP. i just dont understand why its balanced that way, because when I did the balancing, I did not need to add the 6 in front of the irons! PLEASE HELP!!!

Answer 1

6Fe^2+ + 14H^+ + Cr2O7^2- ----> 6Fe^3+ + 2Cr^3+ + 7H2O

Given that

sample weighing 0.5562 g

28.72 mL of 0.04021 M K2Cr2O7 (aq).

Number of mole = molarity * volume in L

= 0.04021* 28.72 ml*1 L/1000 ml

= 0.00115 moles of Cr2O7^2-

Now calculate the moles of Fe2+ which are reacted with these moles of Cr2O7^2-:

0.00115 moles of Cr2O7^2-* 6 mole Fe2+ /1 moles of Cr2O7^2-

= 0.0069 mole Fe2+

Amount in g = number of moles * molar mass

=0.0069 mole Fe2+*55.847 g Fe/ mole

= 0.3853 g Fe 2+

% of Fe2+ = amount of Fe2+/ sample mass*100

=   0.3853 g Fe 2+/0.5562 g *100

= 69.27%

= 69.3%