Asolution of sugar contains 35 gramsof sucrose, c12h22o11in 100 ml of water. what is the percent composition of the solution?

Percent composition of the solution is 26 % of sucrose and 74 % of water

Explanation:

Percent composition is the mass of solute, either of solvent in 100 g of solution.

Mass of solution = Mass of solvent + Mass of solute

Mass of solute = 35 g

Mass of solvent = 100 g

As we know, water density = 1g/mL

So 1g/mL . 100 mL = 100 g

35 g + 100 g = 135 g → Mass of solution

(Mass of solute / Mass of solution) . 100 =

(35 g / 135 g) . 100 = 26 %

(Mass of solvent / Mass of solution) . 100 =

(100 g / 135 g) . 100 = 74 %

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