A chemist reacted 57.50 grams of sodium metal with an excess amount of chlorine gas. The chemical reaction that occurred is shown. Na + Cl2 → NaCl

If the percentage yield of the reaction is 86%, what is the actual yield? Show your work, including the use of stoichiometric calculations and conversion factors.

Explanation:

Moles of Na used

= 57.50g / (22.99g/mol) = 2.51mol.

2Na + Cl2 => 2NaCl

Moles of NaCl = Moles of Na = 2.51mol

Expected mass of NaCl produced

= 2.51mol * (58.44g/mol) = 146.68g.

Since percentage yield is 86%,

Actual mass of NaCl produced

= 146.68g * 0.86 = 126.14g.

94.45 trust me! that other guy is wrong. if you want to find out why go to explanation

Explanation:

To find actual yield:

multiply theoretical yield given by the percentage: 57.50 x 86 = 4945

then divide that by 100: 4945/100 = 49.45

To make sure the its right:

divide the actual yield by the theoretical yield: 49.45/57.50 = 0.86

Then multiply that by 100: 0.86 x 100 = 86%

Thats how you know you got your answer right

There you go :)

Thats how you do it!

hi my dude or dudette