Calculate the number of grams of sodium in 8.4g of NaC7H5O2 (sodium benzoate) express your answer using two significant figures. PLEASE HELP!

1.3g of Na

Explanation:

We'll begin by calculating the molar mass of sodium benzoate, NaC7H5O2. This is shown below:

Molar Mass of NaC7H5O2 = 23 + (12x7) + (5x1) + (17x2)

= 23 + 84 + 5 + 32

= 144g/mol

Now, we can obtain the mass of sodium, Na present in 8.4g of sodium benzoate, NaC7H5O2 as follow:

There are 23g of Na in 144g of NaC7H5O2.

Therefore, the mass of Na in 8.4g of NaC7H5O2 is given by:

23/144 x 8.4 = 1.3g.

Therefore, 1.3g of Na is present in 8.4g of sodium benzoate, NaC7H5O2.

first, consult an activity series chart to determine if al is more reactive than fe. it is, so it will replace the fe. now, one can continue with the problem.

next, make sure you have a correctly balanced equation.

3feo + 2al ==> 3fe + al2o3

now, determine which reactant (feo or al) is in limiting supply.

moles feo present = 125 g x 1 mole/71.8 g = 1.74 moles

moles al present = 25.0 g x 1 mole/26.98 g = 0.927 moles

since from the balanced equation, it take 3 mol feo for every 2 mol al, al will be limiting.

you can determine this simply by dividing moles of each by their coefficient in the balanced equation. for example: 1.74/3 = 0.58 and 0.927/2 = 0.464. al is less, so it is limiting.

finally, using the stoichiometry in the balanced eq. determine mass of fe that can be produced from 0.927 moles of al.

0.927 moles al x 3 moles fe/2 moles al x 55.8 g/mole = 77.6 g

answer; talc and gypsum;