Can someone please answer these chemistry problems about stoichiometry? please?!?

Question

1. A side reaction in the manufacture of rayon from wood pulp is 3CS2 + 6NaOH => 2Na2CS3 + Na2CO3 + 3H2O. How many grams of Na2CS3 are produced in the reaction of 92.5 ml of liq. CO2 (1.26g/ml) and 2.70 moles NaOH?

2. how will you determine the limiting reagent and the excess? and how to compute them?

3. calculate the theoretical yield, potassium chloride produced from the reaction of 2.56 g of K and 3.85 g Cl2. If an actual experiment produces 3.81 g KCl, calculate the percentage yield.

please, can anyone answer these? i really need it, thanks!

Answer 1

1.  I trust you mean CS2, not CO2.  First, you need to find the moles of CS2.  The approach, given the data is

Moles = 92.5 ml x 1.26 g/ml x 1 mole/76 g= 1.5 moles (appx)

The reaction tells us that the reaction uses 1 mole of CS2 per 2 moles of NaOH.   We are given 2.7 moles of NaOH.  Thus, only 1.35 moles of CS2 reacts with it (there is no additional NaOH available to react with the rest).   Now, the reaction tells us that for each 3 moles of NaOH, 1 mole of product Na2CS3 is formed.  Since we have 2.7 moles of NaOH, 0.9 moles of product are formed.  You can multiply 0.9 moles times it mol wt to get your answer.

2.  The above problem gives an example of limiting and excess reagent.  NaOH was limiting because there was not sufficient moles of it to react with the other reagent.  The other reagent is in excess.  

3. Write the rxn:  2K +Cl2 -> 2KCl

Now find the moles of K and moles of Cl2.  

Determine the number of moles of KCl that can be produced with the moles of reactant present (the limiting reagent routine again).   Now find the moles corresponding to 3.81 g KCl.

The percentage yeild is 100 x actual moles/possible moles given reagent mass.