What is the concentration of a solution with a volume of 1.38 mL that contains 17.36 grams of ammonium sulfite [(NH4)2SO3)?

C = 107.97 mol/L

Explanation:

Given data:

Volume of solution = 1.38 mL (1.38 mL× 1 L /1000 mL = 0.00138 L)

Mass of ammonium sulfite = 17.36 g

Concentration of solution =?

Solution:

We will calculate the number of moles of ammonium sulfite.

Number of moles = mass/molar mass

Number of moles = 17.36 g / 116.15 g/mol

Number of moles = 0.149 mol

Concentration:

C = n/V

C = concentration

n = number of moles of solute

v = volume in L

C = 0.149 mol / 0.00138 L

C = 107.97 mol/L

true

explanation:

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