You have a solution of NaOH that is 3.0 M. If you need 250 mL of a 1.5 M solution, what volume of the stock solution should you use?

Explanation: It's a 3.0 M solution so 1 litre of solution contains 3 moles of NaOH. 250 ml of solution therefore contains 0.25 x 3 = 0.75 moles of NaOH. The molar mass of NaOH is 39.9971 g/mol, so 250 ml of this solution contains 0.75 x 39.9971 = 29.99 g, or if you round it up 30.0 g.

Explanation:

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answer is 30 g

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

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