Question

Answer the question in below. Answer all 4 parts, and please show your calculation, and explain a bit.

1. For the hydrolysis of sulfamic acid NH,SO3H (MW 97.1 g.mol-') reacting with Ba(NO3) (MW 261.34 g.mol-') to produce BaSO4 (MW 233.38 g.mol"), we use a solution containing 1.50 g Ba(NO3)2 and 2.50 g of NH2SO3H: Ba(NO3)2(aq) + NH2SO3H(aq) + H20 (1) - BaSO4(s) + NH4NO3(aq) + HNO3(aq) (a) Use the expression below and calculate the grams of sulfamic acid needed to react with 1.50 g of Ba(NO3)2 : grams NH2SO3H = 1.50grams of BaNO3 x mol BaNO3 mol NH2SO3H 8 BaNO3 mol BaNO3 8 NH2SO3H mol NH2SO3H g NH2SO3H needed to react (b) Identify how many grams of sulfamic acid you actually have and see if you have more or less sulfamic acid than the amount you need. If you have more than you need, then BaNO3 is the limiting reactant. If you do not have enough, then sulfamic acid should be the limiting reactant. (c) Assuming BaNO3 is the limiting reactant, calculate the grams of BaSO4 produced using the following expression: g BaSO4 1.5 grams of BaNO3 = mol BaNO3 g BaNO3 mol BaSO4 mol BaNO, mol Baso, g BaSO, produced (d) Knowing that the chemical reaction produces 1.0 g of Baso. Calculate the % yield of the reaction.

Answer 1

0 Theoritical Balanced equation. Inol Ba(NO3), 199+ NH, SO H 199) + H2014 —Bason 6) + Imol I mol NO HYNO 69) + H2O (99 molar masses Bacnos), = 261,34 g/mol NH2SO3H 297.1 glmo ) Basone 233 38 g/mol Theoritical masses Ba(NO3), - 261.34 g NH2SO3 H = 97.19 Basoy 2 233. 38 g The critical moles involved in the reaction is BALNO3), a Imol, NHOSOH = 92.1 mo) Basole : I mol actual massed involved Ba(NO3)2 = 1,50 g NH2 50zH = 2.509

@ Grams of Imol. NHasaH Imol NasaH x Bacnos), 31 = 150g of Ba(NO), 261+34 9 I mol BalNgh of Balno), 97,1g of NH2 SOH I mol H2 SOZ H 97/ X = 0.557 g of wha sozt needed be actually we have 2.50 g of Hy sgt 2.50g-0.557 g = 1.94g ob NHJEH (excess) BacNQ) is limiting reagent. so 9 Im . - sahip) imol 233.38 got 1.50g X 2x Basoy x Basoy - 261.349 Bal Nos? BalN03) Imol Baling) I mol Basoy = 1.34 g of Bascy produe d.

1d Theoritical yield = 1.34 g of Basoy actual yield = 1.0g of Body to yield of Basou z actual yield - Theoritical yield - X 100 1.0 g -x100 2 1,34 g 74.6 % = 75 of