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solmaris [256]
3 years ago
11

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

SAT
1 answer:
siniylev [52]3 years ago
5 0

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 <em>V</em> such that

(0.0198 mol) / <em>V</em> = 0.200 M

Solve for <em>V</em> to get

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

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