a 0.4617g sample containing sodium bicarbonate was dissolved and titrated with standard 0.106M of hydrochloric acid solution , requiring 40.72ml. the reaction is:

HCO3- + H+ -> H2O + CO2

Calculate the percent sodium bicarbonate in the sample. (Given molecular weight of Na2CO3 is 105.99 and HCL is 36.5)

Someone help me please,urgent T-T

I'm confused. You have titrated NaHCO3 but you give the molar mass of Na2CO3. I will assume the 0.4617 g sample did not contain Na2CO3 (or any other carbonate)

mols HCl = M x L = ?
mols HCO3^- = mols HCl since the ratio in the equation is 1:1
Convert mols HCO3^- to grams NaHCO3. g = mols x molar mass NaHCO3.
Then %NaHCO3 = (g NaHCO3/mass sample)*100 = ?

To calculate the percent of sodium bicarbonate in the sample, we need to determine the number of moles of sodium bicarbonate (NaHCO3) present in the sample and then divide that by the total mass of the sample.

First, let's calculate the number of moles of hydrochloric acid (HCl) used in the titration:

Number of moles of HCl = concentration of HCl * volume of HCl used
= 0.106 M * 40.72 ml
= 4.31 mmol

Since the reaction follows a 1:1 stoichiometry between HCl and NaHCO3, the number of moles of HCl used is equal to the number of moles of NaHCO3 present in the sample.

Next, let's calculate the number of moles of NaHCO3:

Number of moles of NaHCO3 = 4.31 mmol

The molar mass of NaHCO3 = 1 * atomic mass of Na + 1 * atomic mass of H + 1 * atomic mass of C + 3 * atomic mass of O
= 23 + 1 + 12 + (3 * 16)
= 23 + 1 + 12 + 48
= 84 g/mol

Now, let's calculate the mass of sodium bicarbonate present in the sample:

Mass of NaHCO3 = number of moles of NaHCO3 * molar mass of NaHCO3
= 4.31 mmol * 84 g/mol
= 362.04 mg

Finally, let's calculate the percent of sodium bicarbonate in the sample:

Percent NaHCO3 = (mass of NaHCO3 / mass of the sample) * 100
= (362.04 mg / 461.7 mg) * 100
≈ 78.50%

Therefore, the percent of sodium bicarbonate in the sample is approximately 78.50%.

To calculate the percent sodium bicarbonate in the sample, we need to determine the number of moles of sodium bicarbonate reacted with hydrochloric acid.

First, we need to find the number of moles of HCl used in the reaction.

In this case, the molarity (M) of the hydrochloric acid solution is given as 0.106 M. The volume (V) of HCl used is 40.72 mL, which can be converted to liters by dividing by 1000: 40.72 mL/1000 = 0.04072 L.

Using the formula:

moles of HCl = Molarity (M) × Volume (L)

we can calculate the moles of HCl:

moles of HCl = 0.106 M × 0.04072 L = 0.00431592 moles of HCl

According to the balanced equation, we see that one mole of HCl reacts with one mole of sodium bicarbonate (NaHCO₃). Therefore, the number of moles of sodium bicarbonate can be determined by the mole ratios:

moles of NaHCO₃ = 0.00431592 moles of HCl

Next, we can determine the mass of sodium bicarbonate using the given molecular weight (105.99 g/mol):

mass of NaHCO₃ = moles of NaHCO₃ × molecular weight

mass of NaHCO₃ = 0.00431592 moles of NaHCO₃ × 105.99 g/mol = 0.45731 g

Finally, we can calculate the percent sodium bicarbonate in the sample:

percent sodium bicarbonate = (mass of NaHCO₃ / mass of sample) × 100%

percent sodium bicarbonate = (0.45731 g / 0.4617 g) × 100% = 99.04%

Therefore, the percent sodium bicarbonate in the sample is approximately 99.04%.