Question

SCl2 +NaF --> S2Cl2 +SF4 + NaCl

1, balance equation

2. How many moles of NaCl does this reaction prouce based upon the reation of 7.34g of NaF with an excess of SCl2?

3. How many grams of NaCl are formed?

4. What is the percent yeild if your experimental yeild is 9.22g of NaCl

Answer 1

Ans 1 :

To balance the reaction , identify the elements on each side and note it down for your convenience. Then equalise the number of atoms on each side of the reaction.

The balanced reaction is given as :

3SCl2 + 4NaF = S2Cl2 + SF4 + 4NaCl

Ans 2 :

number of moles of NaF = 7.34 / 41.99 = 0.175 moles

So the number of moles of NaCl produced will also be 0.175 moles

Ans 3 :

Mass of NaCl produced = 0.175 x 58.44

= 10.23 grams

Ans 4 :

Percent yield = ( actual yield / theoretical yield) x 100

= (9.22 / 10.23) x 100

= 90.13 %