The fizz produced when a tablet of alka-seltzer dissolves in h2o comes from the reaction of sodium carbonate,, nahco3, and citric acid, c6h8o7: . 3nahco3+c6h8o7--> 3co2+3h2o+na3c6h5o7. supposed the tablet containing 1.00g of sodium bicarbonate and 1.00g of citric acid reacts to completion. how many grams of co2 are produced and how many grams of the excess reagent remain when the reaction is complete?

1st step.  Figure out the limiting reagent.  Calculate moles (mass/molar mass).

moles of sodium bicarbonate = 1/84 = 0.0119

moles of citric acid = 1/192 = 0.0052

2nd step. Divide moles by the coefficient in the equation to see which is limiting:

Sodium bicarbonate: 0.0119/3 = 0.0040

Citric acid: 0.0052/1 = 0.0052

Based on these conclusions, sodium bicarbonate is the limiting reagent.

Since the coefficients of sodium bicarbonate and CO2 are the same in the equation, the number of moles of sodium bicarbonate consumed equals the number of moles of CO2 produced. Therefore 0.0119 moles of CO2 are produced, which is 0.524 g (after multiplying by molar mass).

Given: 3 moles of sodium bicarbonate react with 1 mole of citric acid. 

It means you have a third as many moles of citric acid as sodium bicarbonate, which in this case is 0.119/3 = 0.0040 moles. However, we know from above that there were 0.0052 moles to begin with. Therefore, after the reaction, there are 0.0052-0.0040=0.0012 moles of citric acid remaining. Multiplying by the molar mass of citric acid, we get 0.238 g remaining.

when [oh-] is 10-5  m, and [h3o+] is 10-9  m, the poh is 5 and ph is 9. again in both cases the sum of ph and poh is 14. however, the hydrogen ion concentration is not always going to be equal to exactly 1 x 10 raised to a negative number. for example, we skipped over the value of 2.0 x 10-7.