Two More questions?

The 3rd one is balanced and correct, not the second one. THe second equation has 8 Cl atoms on the left,and only 2 on the right.

4 mol Cl2 X 3 molFeCl3/3 mol Cl2 X _____ grams FeCl3/mol (I'll let you calculate the molecular weight of FeCl3)

In the first part, the first and second equations all have coefficients that can be divided by 2, so they are not correct. The fourth one is not balanced. This leaves the third one.

2Na + 2H2O -> 2NaOH + H2

In the second part, the correct equation is:

2FeI2 + 3Cl2 -> 2FeCl3 + 2I2

3 moles of Cl2 form 2 moles of FeCl3, so 4 moles of Cl2 form (2/3)×4 moles of FeCl3. The molar mass of FeCl3 is 162.2, so this works out as [(2/3)×4]×162.2 = 432.53 grams FeCl3.

The answer is 2Na+2H2O=2NaOH+H2

On the left hand side there are

2 Na

4 H (2 x H2)

2 O (2 x O)

and on the right hand side

2 Na

4 H (2 H in NaOH and 2 in H2)

2 O (2 O in NaOH)

So, both sides equal and it's the smallest number.

The second eqn is wrong since

Left hand side has 2 Na and 8 Cl ( 4 x Cl2)

Right hand side has 2 Na and 2 Cl

So not balanced!!!

To determine the number of grams of FeCl3 from 4.00 moles of Cl2, write:

2Fe+3Cl2 --> 2FeCl3

then, we see that 3 x no. of moles of Cl2 = 2 x no. of moles of FeCl3

therefore, no. of moles of FeCl3 is 6

No. of grams=no. of moles x Molecular mass of FeCl3

= 6 x (56+3x35.5)

= 975 gram