Baking powder must contain 1 mole of potassium bitartrate (KNaC4H4O6) for every mole of sodium bicarbonate (NaHCO3). If you have 1 mole of NaHCO3, calculate how many grams of KNaC4H4O6 you will need.

Answer 1

Hello, Jaydon,

We are given 1 mole of NaHCO3 and are told we need 1 mole of KNaC4H4O6 per mole of the sodium bicarbonate. Therefore we need 1 mole of the potassium bitartrate. The answer is requested in grams, so we need the molar mass of the bitartrate. I calculate 210.13 g/mole.

We need 213.13 grams.

Bob

Answer 2

The compound you present is not potassium bitartrate, but is sodium potassium tartrate. Potassium bitartrate would be KC₄H₅O₆. Since you typed KNaC4H4O6 TWICE in your question, I'll assume you mean sodium potassium tartrate.

1 mol NaHCO₃ will have 1 mol KNaC₄H₄O₆

molar mass KNaC₄H₄O₆ = 210 g/mol

So 1 mol of NaHCO₃ would contain 210 g of KNaC₄H₄O₆