Mass of Ba(NO3)2= 1.3550 grams Mass of NH2SO3H= 2.4934 grams Mass of precipitate
ID: 1017704 • Letter: M
Question
Mass of Ba(NO3)2= 1.3550 grams
Mass of NH2SO3H= 2.4934 grams
Mass of precipitate= 1.2798 grams
Standard Diviation (mass of precipitate/mass of limiting reactant): 0.2517
Our tables ratio (mass of precipitate/mass of limiting reactant): 0.9423
mean value of the ratio that is consistant with the precision of the data: 0.7401
1.c : mass for the precipitate (mean value of the ratio & mass of your limiting reactant): 1.0028g (????)
2.a: calculate the expected masses of the 3 possible products. Use the mass of the limiting reactant and the equations that you balanced in the prelab assignment
Ba(NH2SO3)2:
BaSO4:
Ba(NH2)2:
b. compare the mass from Q 1.C with the masses obtained in Q 2.A. Judging on the basis of the comparison, what is the identiry of the precipitate?
Limiting reactant is Ba(NO3)2
balanced equations are:
Ba(NO3)2+ 2 NH2SO3H+ 0 H2O ---> 1 Ba(NH2SO3)2+ 2 HNO3
Ba(NO3)2+ 1 NH2SO3H+ 1 H2O ---> 1 BaSO4+ 1 NH4NO3+ 1 HNO3
Ba(NO3)2+ 2 NH2SO3H+ 2 H2O --->1 Ba(NH2)2 + 2 H2SO4+ 2 HNO3
Explanation / Answer
2 a) The limiting reactant in all three equations is Ba(NO3)2
- Mol wt of Ba(NO3)2 = 261.337g/mol
- # moles of Ba(NO3)2= 1.3550/261.337 = 0.00518 moles
Since the amount of the precipitate formed depends on the amount of the limiting reagent, we can calculate the expected masses of the products as follows:
- Mol wt of Ba(NH2SO3)2 = 329.49 g/mol
Expected mass of Ba(NH2SO3)2 = 0.00518 * 329.49 = 1.7067 g
- Mol wt of BaSO4 = 233.43 g/mol
Expected mass of BaSO4 = 0.00518 * 233.43 = 1.2092 g
- Mol wt of Ba(NH2)2 = 169.37 g/mol
Expected mass of Ba(NH2)2 = 0.00518 * 169.37 = 0.8773 g
2 b) The mass of the precipate obtained from the experiment = 1.2798 g which is close to the expected value of 1.2092 g. therefore, the identity of the precpitate is BaSO4.