To yield a total of 6 grams sodium citrate, how many grams of sodium bicarbonate (baking soda) will you need?
MM = molar mass
3NaHCO3 + zzzz ==> Na3citrate
6 g x (MM Na3citrate/3*MM NaHCO3) = g NaHCO3 need to make 6 g sodium citrate assuming the yield is 100%.
:/ (i am not a real teacher)
i dont take this class
To calculate the amount of sodium bicarbonate (baking soda) needed to yield a total of 6 grams of sodium citrate, we need to consider their respective molecular weights. The molecular weight of sodium citrate (Na3C6H5O7) is approximately 294.10 g/mol, while the molecular weight of sodium bicarbonate (NaHCO3) is approximately 84.01 g/mol.
To find the amount of sodium bicarbonate needed, we can set up a ratio using the molecular weights:
(Molecular weight of sodium citrate / Molecular weight of sodium bicarbonate) = (Amount of sodium citrate / Amount of sodium bicarbonate)
(294.10 g/mol / 84.01 g/mol) = (6 g / x)
Simplifying the ratio:
3.4988 = (6 g / x)
Cross-multiplying:
3.4988x = 6 g
Dividing by 3.4988:
x = 6 g / 3.4988
Therefore, you will need approximately 1.71 grams of sodium bicarbonate (baking soda) to yield a total of 6 grams of sodium citrate.
To determine the amount of sodium bicarbonate (baking soda) needed to yield a total of 6 grams of sodium citrate, we first need to find the molar ratio between sodium citrate and sodium bicarbonate.
The molecular formula of sodium citrate is C6H5Na3O7, and the molecular formula of sodium bicarbonate is NaHCO3. By comparing the two formulas, we can see that there is one sodium ion (Na+) in sodium bicarbonate and three sodium ions (3Na+) in sodium citrate.
We can use this information to calculate the molar ratio between sodium citrate and sodium bicarbonate. Since the molar mass of sodium citrate is 258.07 grams/mol and the molar mass of sodium bicarbonate is 84.01 grams/mol, the ratio can be expressed as:
(3 mol of Na+ in sodium citrate) / (1 mol of Na+ in sodium bicarbonate)
To calculate the amount of sodium bicarbonate needed, we can set up a proportion using the molar ratio:
(3 mol of Na+ in sodium citrate) / (1 mol of Na+ in sodium bicarbonate) = (6 grams of sodium citrate) / (x grams of sodium bicarbonate)
Cross-multiplying and solving for x gives:
3 mol of Na+ / 1 mol of Na+ = 6 grams of sodium citrate / x grams of sodium bicarbonate
3/x = 6/84.01
Cross-multiplying and solving for x:
3 * 84.01 = 6 * x
252.03 = 6x
x = 252.03 / 6
x ≈ 42 grams
Therefore, to yield a total of 6 grams of sodium citrate, you would need approximately 42 grams of sodium bicarbonate.