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!

the answer is definitely c.

when the volume of gas is   changed from   717.73 ml   to 852 ml   the temperature will   change   from 315   c to 452   c

    calculation

this is calculated using the charles law formula

that is,   v1/t1 = v2/t2

          v1=?

        t1 = 315 c   into   kelvin = 315 +273 = 588 k

        v2 = 852   ml

        t2 = 425c + 273 =698 k

make v1   the subject of the formula ( by   multiplying both side by t1)

  v1=( v2 xt1) / t2

v1= (852 ml   x 588 k)/ 698 k= 717.73 ml