Question

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.

Answer 1

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 = ?%

Answer 2

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