Asked by Studying
13.In a study of rust-removing solutions, 27.8 mL of 0.115 mol/L phosphoric acid reacts completely with 0.245 mol/L sodium hydroxide. Predict the minimum volume of sodium hydroxide required for this reaction?
3 NaOH + H3Po4 = 3H20 + Na3PO4
(27.8 ml)(0.115) = 0.003197 mol H3Po4
0.003197 * (3 mol NaOh) = 0.009591 mol
v = 0.009591 mol/ 0.245 mol/L
v = 39.1 ml
3 NaOH + H3Po4 = 3H20 + Na3PO4
(27.8 ml)(0.115) = 0.003197 mol H3Po4
0.003197 * (3 mol NaOh) = 0.009591 mol
v = 0.009591 mol/ 0.245 mol/L
v = 39.1 ml
Answers
Answered by
DrBob222
You made two error in naming but they cancel.
27.8 x 0.115 is 0.003197 MILLImoles (not mols) BUT
at the end you divide millimoles/mL and that is M. Your answer of 39.1 mL is right.
27.8 x 0.115 is 0.003197 MILLImoles (not mols) BUT
at the end you divide millimoles/mL and that is M. Your answer of 39.1 mL is right.
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