How many milligrams of NaHCO3 are in a 500mg tablet if 40.0 mL of 0.120M HCL is required to neutralize the sample?
This is very like the KHP question I have just done for you.
1.write out the equation for the reaction.
2. calculate the number of moles of HCl used.
3. from the equation deduce the number of moles of NaHCO3 there must have been.
4. calculate the molar mass of NaHCO3
5. use the result of steps 3 and 4 to calculate the mass of NaHCO3.
6. convert result to milligrams.
i got 0.01371516348 moles of HCl and then the molecular mass of NaHCO3 is 84.008..
how am i going to use these 2 to find the mass of NaHCO3?
(sorry, i'm just confused)
First of all 0.01371516348 is a silly number of significant figures to use as the starting data is only to 3 sig figs.
Haveyou written out the starting equation, step 1?
The number of moles of HCl you can do in your head really
0.040 L x 0.120 mole L^-1=
i didn't round it off, because in our class we have a rule that we should not round off until we get the final answer.
NaHCO3 + HCl -> NaCl + H2O + o2
0.040 L / 0.120 mole L = 0.33..
oh, you used the molarity to get the mole..
but i don't know what will i do with the 500mg
and how to relate (3)and(4) to get the mass of NaHCO3.
how do you do step 3?
To find out how many milligrams of NaHCO3 are in a 500mg tablet, we need to use the concept of stoichiometry.
First, we start by converting the volume of HCl solution to moles of HCl using the given concentration (0.120M) and volume (40.0 mL).
Using the formula:
moles of solute = concentration × volume (in liters)
First, we convert milliliters to liters:
40.0 mL ÷ 1000 = 0.04 L
Now we can find the number of moles of HCl:
moles of HCl = 0.120M × 0.04 L = 0.0048 moles
Next, we use the balanced chemical equation for the reaction between HCl and NaHCO3:
2 HCl + NaHCO3 -> NaCl + H2O + CO2
From the equation, we can see that the stoichiometric ratio between HCl and NaHCO3 is 2:1. Meaning, for every 2 moles of HCl, we need 1 mole of NaHCO3.
Since we have 0.0048 moles of HCl, we can determine the number of moles of NaHCO3 needed:
moles of NaHCO3 = 0.0048 moles ÷ 2 = 0.0024 moles
Lastly, we can determine the mass of NaHCO3 in milligrams using its molar mass:
molar mass of NaHCO3 = 23 g/mol (Na) + 1 g/mol (H) + 12 g/mol (C) + 3 x 16 g/mol (O) = 84 g/mol
mass of NaHCO3 = 0.0024 moles × 84 g/mol = 0.2016 grams
Converting to milligrams:
mass of NaHCO3 = 0.2016 grams × 1000 = 201.6 milligrams
Therefore, there are approximately 201.6 milligrams of NaHCO3 in the 500mg tablet.