Question

how many milliliters of 0.200 m nh₄oh are needed to react with 12.0 ml of 0.550 m fecl₃? fecl₃+ 3nh₄oh → fe(oh)₃ + 3nh₄cl

Answer 1

You're starting with

(12.0 mL) (0.550 M) = (0.0120 L) (0.550 M) = 0.00660 mol

of FeCl₃.

For every mole of FeCl₃ in the balanced reaction, you need 3 moles of NH₄OH, so you need a minimum 0.0198 mol NH₄OH per mole FeCl₃.

Then the requisite volume of NH₄OH solution needed is V such that

(0.0198 mol) / V = 0.200 M

Solve for V to get

V = (0.0198 mol) / (0.200 M) = 0.0990 L = 99 mL