Calling All Chemistry Geniuses!!!! HELP!!?

[John-G]

In the following reaction is run by combining 95.0g of NaOH, 141g of Cl2, and 125g of Ca(OH)2.

2NaOH + Ca(OH)2 + 2Cl2-------> Ca(OCl)2 + 2NaCl +2H20

Determine:

(a) The Limiting Reagent.


(b) How mcuh is left over of EACH reagent.


(c) The amount of Ca(OCl)2 that can be made.


(d) The % yield if the amount of water collected was 31.5g.






Thanks so much for your time and consideration it is GREATLY appreciated!!!

[cricketluvr]

(a) use stoichiometry to determine this

start with either reactant...let's jhust go with NaOH

95 g NaOH * (1 mole NaOH / 40 g) * (1 mole Ca(OH)2/ 2 mole NaOH) * (74 g Ca(OH)2 / 1 mole Ca(OH)2)=87.88 g Ca(OH)2

We start out with 125 g Ca(OH)2, so there is enough of that. The limiting reagent now is either Cl2 or NaOH.

95 g NaOH * (1 mole NaOH / 40 g) * (2 mole Cl2 / 2 mole NaOH) * (71 g Cl2 / 1 mole Cl2 =166.3 g Cl2

We do not have that much Cl2 to work with, so Cl2 is the limiting reagent.




(b) for this, use stoichiometry, and begin with Cl2 because it is the limiting reagent.

141 g Cl2 * (1 mole Cl2 / 71 g Cl2) * (1 mole Ca(OH)2 / 2 mole Cl2) * (74 g Ca(OH)2 / 1 mole Ca(OH)2) = 73.5 g Ca(OH)2 used

141 g Cl2 * (1 mole Cl2 / 71 g Cl2) * (2 mole NaOH / 2 mole Cl2) * (40 g NaOH / 1 mole NaOH) = 79.4 g NaOH used

since we have 125 g Ca(OH)2 to begin with, we are left with 51.5 g Ca(OH)2, and since we have 95 g NaOH to begin with, we are left with 15.6 g NaOH. The Cl2 is completely used up because it is the limiting reagent.



(c) you start out with Cl2 again because this is the limiting reagent.

141 g Cl2 * (1 mole Cl2 / 71 g Cl2) * (1 mole Ca(OCl)2 / 2 mole Cl2) * (143 g Ca(OCl)2 / 1 mole Ca(OH)2) = 283.99 g Ca(OCl)2 produced




(d) first find out how much water will be produced, and then place the actual yield over theoretical yield.

141 g Cl2 * (1 mole Cl2 / 71 g Cl2) * (2 mole H2O / 2 mole Cl2) * (18 g H2O / 1 mole H2O) = 35.75 g H2O produced

% yield = (31.5 / 35.75) * 100 = 88.12%