A 0.484 g sample of impure NH4Br is treated with 25.00 mL of 0.2050 M NaOH & heated to drive off the NH3. The unreacted NaOH in the reaction mixture after heating required 9.95 mL of 0.0525 M H2C2O4 to neutralize. How many grams of NH4Br were in the original sample? What is the percentage NH4Br in the original sample?

I don't even know where to start. Please help with the steps.

Here is what's going on.

NH4Br + NaOH + heat ==> NH3(g) + HBr(g)
How many mols NaOH did you add? That's M x L = 0.025 L x 0.2050M = approximately 0.0051 but you need to do it more accurately. This 0.0051 mols is more NaOH that is needed so the excess is titrated with H2C2O4 (oxalic acid).

H2C2O4 + 2NaOH ==> 2H2O + Na2C2O4
How many mols H2C2O4 were used? That's M x L = 0.00995L x 0.0525M = 0.00052. Convert mols H2C2O4 to mols NaOH.
0.00052 x (2 mols NaOH/1 mol H2C2O4) = about 0.001. That's how much too much NaOH you had initially.
0.0051 initially - 0.001 mol NaOH left = 0.04 mols NaOH used in the NH4Br treatment.
Since 1 mol NaOH = 1 mol NH4B4, then grams NH4Br = mols NaOH used x molar mass NH4Br. That gives you the grams pure NH4Br in the sample.
Then %NH4Br = (g NH4Br/mass sample)*100 = ?%

I don't really understand which is the mass of the sample.

Well, this problem seems to have a lot of letters and numbers hanging out together. It's like a math party! Don't worry, I'll break it down for you.

Step 1: Let's figure out the number of moles of NaOH used in the reaction.

To find the moles of NaOH, we use the formula:

moles = concentration (in M) × volume (in L)

So, moles of NaOH = 0.2050 M × 0.02500 L

Step 2: Now, let's find the moles of H2C2O4 used to neutralize the leftover NaOH.

Using the same formula as above, we have:

moles of H2C2O4 = 0.0525 M × 0.00995 L

Step 3: The moles of H2C2O4 reacted with the unreacted NaOH are equal to the moles of NaOH in the reaction.

So, moles of H2C2O4 = moles of NaOH

Step 4: The moles of H2C2O4 can be used to find the moles of NaOH that reacted with NH4Br.

Since the reaction between NaOH and H2C2O4 is 1:2, the moles of NaOH that reacted with NH4Br are half the moles of H2C2O4 reacted.

moles of NaOH (reacted with NH4Br) = moles of H2C2O4 / 2

Step 5: Now, let's find the moles of NH4Br.

Since the reaction between NH4Br and NaOH is 1:1, the moles of NH4Br are equal to the moles of NaOH that reacted with NH4Br.

moles of NH4Br = moles of NaOH (reacted with NH4Br)

Step 6: Finally, let's find the mass of NH4Br.

To find the mass of NH4Br, we use the formula:

mass = moles × molar mass

The molar mass of NH4Br is 99.45 g/mol.

mass of NH4Br = moles of NH4Br × molar mass

And there you have it! That's the step-by-step process to solve this problem. Just plug in the numbers and do the math to find the grams of NH4Br in the original sample and the percentage of NH4Br in the sample. Remember, math problems are like puzzles, and solving them is like putting together the pieces of a clown outfit. Good luck!

To find out the number of grams of NH4Br in the original sample and the percentage of NH4Br in the sample, we need to use stoichiometry and titration calculations.

Let's break down the problem into steps:

Step 1: Calculate the number of moles of NaOH used in the reaction and the number of moles of unreacted NaOH after heating.
- Given: Volume of NaOH solution = 25.00 mL = 0.02500 L
Concentration of NaOH solution = 0.2050 M
- Use the formula: moles = volume (L) x concentration (M)
- Therefore, moles of NaOH used = 0.02500 L x 0.2050 M = 0.005125 mol

Step 2: Calculate the number of moles of H2C2O4 consumed to neutralize the unreacted NaOH after heating.
- Given: Volume of H2C2O4 solution = 9.95 mL = 0.00995 L
Concentration of H2C2O4 solution = 0.0525 M
- Use the formula: moles = volume (L) x concentration (M)
- Therefore, moles of H2C2O4 consumed = 0.00995 L x 0.0525 M = 0.000521 mol

Step 3: Determine the ratio between NaOH and NH4Br in the reaction equation.
- The balanced equation for the reaction between NaOH and NH4Br is as follows:
NaOH + NH4Br -> NaBr + NH3 + H2O
- From the equation, we can see that the ratio between NaOH and NH4Br is 1:1.

Step 4: Calculate the number of moles of NH4Br in the original sample.
- Since the ratio between NaOH and NH4Br is 1:1, the number of moles of NH4Br equals the number of moles of unreacted NaOH.
- Therefore, moles of NH4Br = 0.000521 mol

Step 5: Calculate the mass of NH4Br in the original sample.
- Given: Mass of impure NH4Br sample = 0.484 g
- Use the formula: mass = moles x molar mass
- The molar mass of NH4Br = (1 x molar mass of N) + (4 x molar mass of H) + molar mass of Br.
- The molar masses are: molar mass of N = 14.01 g/mol, molar mass of H = 1.01 g/mol, molar mass of Br = 79.90 g/mol.
- Therefore, mass of NH4Br = 0.000521 mol x [(1 x 14.01 g/mol) + (4 x 1.01 g/mol) + 79.90 g/mol] = 0.0670 g

Step 6: Calculate the percentage of NH4Br in the original sample.
- Use the formula: percentage = (mass of NH4Br / mass of impure sample) x 100
- Therefore, percentage of NH4Br = (0.0670 g / 0.484 g) x 100 = 13.8%

So, there are 0.0670 grams of NH4Br in the original sample, and the percentage of NH4Br in the sample is 13.8%.