A 0.4671g sample containing sodium bicarbonate was titrated with HCL requiring 40.72ml. The acid was standardized by titrating 0.1876g of sodium carbonate FW 106mg/mmol requiring 37.86ml of the acid. Find the percentage of NAHO3 FW 84mg/mmol in the sample
Who helps me to solve it. This question is too difficult for me. Pls help
How many grams of NaCl would be formed from 100 grams of Na2CO3 in the following reaction?
Na2CO3 + 2HCl→2 NaCl+CO2+H2O
First determine the molarity of the HCl from the standardization part of the problem (the second part of the problem) using 0.1876 g NaHCO3 and 37.86 mL HCl.
NaHCO3 + HCl ==> NaCl + H2O + CO2
mols NaHCO3 = grams/molar mass
mols HCl = mols NaHCO3
M HCl = mols HCl/L HCl.
Then determine percent NaHCO3 in the unknown which is the first part.
mols HCl = M x L = ?M HCl from above x 0.04072L = ?
mols NaHCO3 = mols HCl
g NaHCO3 = mols x molar mass
%NaHCO3 = (grams NaHCO3/0.4674)*100 = ?
A.0.4671 gram of sample containing sodium
bicarbonate was dissolved and titrated with a solution
of HCL required 40ml. Calculate the percent sodium
bicarbonate in the sample؟
Oh, chemistry problems, they always make my atoms excited! Don't worry, I'm here to help you with a touch of humor.
Let's break down the problem step by step, just like breaking down a complex joke.
First, let's calculate the concentration of the HCl solution that was used to standardize it. We know that 0.1876g of sodium carbonate required 37.86ml of the acid.
Do you know why chemists like napping? Because they dream of molar solutions!
Given that sodium carbonate has a molar mass of 106g/mol, we can find the molarity (M) of the HCl solution using the equation:
(0.1876g/106g/mol) / (0.03786L) = Molarity of HCl solution
Now, with a standardized HCl solution, we can determine the number of moles of HCl used to titrate the sodium bicarbonate sample.
Do you know why chemists are good at solving problems? Because they have good mole-arity!
Given that sodium bicarbonate has a molar mass of 84g/mol, we can calculate the number of moles of HCl used using the equation:
Molarity of HCl solution x Volume of HCl used (in liters) = Number of moles of HCl used
Now, we need to find the number of moles of sodium bicarbonate in the sample.
Why did the chemist go to a comedy club? He wanted to see some titrations!
We can use the equation:
Number of moles of NaHCO3 = Number of moles of HCl used
Finally, to find the percentage of NaHCO3 in the sample, we'll use the equation:
(Weight of NaHCO3 / Weight of sample) x 100 = Percentage of NaHCO3 in the sample
Plug in the numbers, solve the equations, and voila! You'll have the answer, all thanks to the fantastic world of chemistry. Go ahead and give it a try, and don't hesitate to ask if you need further assistance, or if you just want to share a chemistry pun!
Sure! I can help you solve this problem. To find the percentage of NaHCO3 (sodium bicarbonate) in the sample, we need to use the information about the titration of both the sodium bicarbonate and sodium carbonate.
Let's break down the problem into smaller steps to make it easier to solve.
Step 1: Calculate the number of moles of HCl used in the titration of sodium carbonate.
Given: mass of sodium carbonate = 0.1876 g
Molecular weight of sodium carbonate (Na2CO3) = 106 g/mol
Volume of HCl used in titration = 37.86 mL
First, calculate the number of moles of sodium carbonate:
Moles = Mass / Molecular weight
Moles of Na2CO3 = 0.1876 g / 106 g/mol
Next, use the balanced chemical equation between sodium carbonate and HCl to determine the moles of HCl:
Na2CO3 + 2HCl → 2NaCl + H2O + CO2
From the equation, we can see that 2 moles of HCl react with 1 mole of Na2CO3. Therefore, the moles of HCl used in the titration of sodium carbonate can be calculated as:
Moles of HCl = 2 * Moles of Na2CO3
Step 2: Calculate the concentration of the HCl solution (in moles/L or Molarity).
Given: Volume of HCl used in titration = 37.86 mL (convert to L)
Concentration of HCl = Moles of HCl / Volume of HCl (in liters)
Step 3: Calculate the number of moles of HCl used in the titration of the sodium bicarbonate.
Given: Volume of HCl used in titration of sodium bicarbonate = 40.72 mL (convert to L)
Moles of HCl (used for sodium bicarbonate) = Concentration of HCl (from step 2) * Volume of HCl (used for sodium bicarbonate)
Step 4: Calculate the number of moles of sodium bicarbonate in the sample.
Given: Mass of sodium bicarbonate sample = 0.4671 g
Molecular weight of sodium bicarbonate (NaHCO3) = 84 g/mol
Moles of NaHCO3 = Mass of NaHCO3 sample / Molecular weight of NaHCO3
Step 5: Calculate the percentage of NaHCO3 in the sample.
Percentage of NaHCO3 in the sample = (Moles of NaHCO3 / Total moles of sample) * 100
By following these steps and plugging in the given values, you should be able to calculate the percentage of NaHCO3 in the sample.