Asked by Jarralynn
One 240. mL serving of ChemCola™ contains 57.0 mg phosphorus (which is found in the phosphoric acid, H3PO4, in ChemCola). What volume (in mL) of 0.0950 M NaOH is needed to neutralize the H3PO4 in one serving of ChemCola™?
Answers
Answered by
DrBob222
H3PO4 + 3NaOH ==> Na3PO4 + 3H2O
How many mols in 57 mg of P. Mols = grams/atomic mass = 0.057/approx 31 but you need to redo that and all calculations that follow. I've just estimated all calculations. mols P = 0.00184 mols. That's also 0.00184 mols H3PO4. Now convert mols H3PO4 to mols NaOH.
That's 0.00184 mols H3PO4 x (3 mols NaOH/1 mol H3PO4) = estimated 0.0055 mols NaOH.
Then mols NaOH = M x L = 0.0055 = M x L. You know M from the problem of the NaOH, substitute and solve for L, then convert to mL. Post your work if you get stuck.
How many mols in 57 mg of P. Mols = grams/atomic mass = 0.057/approx 31 but you need to redo that and all calculations that follow. I've just estimated all calculations. mols P = 0.00184 mols. That's also 0.00184 mols H3PO4. Now convert mols H3PO4 to mols NaOH.
That's 0.00184 mols H3PO4 x (3 mols NaOH/1 mol H3PO4) = estimated 0.0055 mols NaOH.
Then mols NaOH = M x L = 0.0055 = M x L. You know M from the problem of the NaOH, substitute and solve for L, then convert to mL. Post your work if you get stuck.
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