ok...this ones kind of a doozy for me?

Question

A decomposition reaction separately occurred for 2 samples (a & b) which contained NaHCO3 (solid) (84 g/mol). Both samples weighed the same but one was pure and one was impure. The product of the decomposition of sample a weighed .25 g and the product of sample b weighed .29 g. Assuming all impurities evaporated during heating:

a. determine the grams of NaHCO3 originally in the pure sample

b determine the grams of impurity in the impure sample

c. determine the % of purity

Answer 1

2NaHCO3(s) --> Na2CO3(s) + CO2(g) + H2O(g)

a) the product weighed 0.25g was Na2CO3

Na2CO3 MW = 105.99 g/mole

0.25 g Na2CO3 = 0.25 g * mole/105.99g = 0.00236 moles

2 NaHCO3 --> Na2CO3 so 2 * 0.00236 mole NaHCO3 =

= 0.00472 moles gave rise to this product

NaHCO3 MW = 84.007 g/mole

0.00472 moleNaHCO3 = 0.00472 * 84.007 g/mole =

= 0.39 g NaHCO3

b) following from example a 0.29 product Na2CO3

0.29 g * mole/105.99g = 0.00274 *2 * 84.007 g/mole =

= 0.46 g NaHCO3

c) If both sample weighed the same

weight impure sample (less weight)/ weight pure sample (more weight) *100

= 0.39/046 * 100 = 85 % pure

Answer 2

I do not thinkyou have enough information to answer this question. You need to know the composition of the decomposition product. If we assume that the decomposition is pure NaHCO3 and the decomposition was 100% effective, then there were 0.25 grams of NaHCO3 to start with. THen the grams of impurity are the difference 0.29-.025=0.04 grams
The %purity would be .25/.29

good luck.