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Consider the reaction H3PO4 + 3NaOH → Na3PO4 + 3H2O. How much Na3PO4

can be prepared by the reaction of 3.43 g of H3PO4 with an excess of NaOH?

For the reaction ?Fe+?H2O → ?Fe3O4 +?H2, what is the maximum amount of Fe3O4(231.533 g/mol) which could be formed from 12.34 mol of Fe (55.845 g/mol) and 8.74 mol of H2O (18.0153g/mol)?

and the last question would be:
What mass of water is needed to react completely with the Na2O2?


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2 answers
  1. The first one is a regular stoichiometry problem and I know one was worked for you yesterday. You should need only one worked of each type because you need to learn to do them yourself.
    Here is a 3/4 step process that will work aboutr 99.9% of all of the stoichiometry problems.
    1. Write and balance the equation.
    2. Convert what you have (in this case H3PO4) to mols. mols = grams/molar mass
    3. Convert what you have (mols H3PO4) to mols of what you want. To do that you use the coefficients in the balanced equation.
    4. Then you convert mols of what you want to grams. grams = mols x molar mass.
    (Note: In step 1 if you are given M and L instead of grams, then mols = M x L = ?. In step 4 if the question asks for volume of a gas, the use PV = nRT to solve for volume._

    #2 is a limiting reagent problem. You know that because amounts are given for BOTH reactants. All LR problems are "extended" stoichiometry problems. They are done this way.
    1. Write and balance the equation. This is the same as above in regular stoichiometry (RS).
    2a. Convert grams of one reagent to mols and that's done the same way.
    2b. You also have another reagent with a value. Convert that to mols as above. Same as RS.
    3a. Now convert mols of what you have (reagent 1) to mols of the product. Same process as RS.
    3b. Do the same for the reagent 2. Same as RS
    3c. Here is the only difference. It is likely that the two values will not agree which means one of them is wrong. In LR problems the correct value is ALWAYS the smaller value ad the reagent responsible for that value is the LR.
    4. Now convert the smaller amount to grams. Same as RS which is g = mols x molar mass

    You should be able to work the last one by yourself; it is a regular stoichiometry problem. . The equation for this reaction is
    2Na2O2 + 2H2O ==> 4NaOH + O2
    Na2O2 + 2H2O ==> 2NaOH + H2O2

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  2. Thank YOU!

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