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?
PLEASE, HELP YOUR FRIEND!!!
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
Thank YOU!
To determine the amount of Na3PO4 that can be prepared in the reaction, we need to use stoichiometry and molar ratios.
Given:
Mass of H3PO4 = 3.43 g
Molar mass of H3PO4 = 98.00 g/mol
First, we need to find the number of moles of H3PO4:
Number of moles = Mass / Molar mass
Number of moles of H3PO4 = 3.43 g / 98.00 g/mol = 0.035 moles
According to the balanced equation:
1 mole of H3PO4 reacts with 1 mole of Na3PO4
Therefore, the number of moles of Na3PO4 that can be prepared is also 0.035 moles.
Now, we need to find the mass of Na3PO4:
Mass = Number of moles × Molar mass
Mass of Na3PO4 = 0.035 moles × 164.00 g/mol = 5.74 g
Hence, the maximum amount of Na3PO4 that can be prepared is 5.74 g.
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For the reaction Fe + H2O → Fe3O4 + H2, we need to use stoichiometry and molar ratios to find the maximum amount of Fe3O4 that can be formed.
Given:
Moles of Fe = 12.34 mol
Molar mass of Fe = 55.845 g/mol
Moles of H2O = 8.74 mol
Molar mass of H2O = 18.0153 g/mol
Molar mass of Fe3O4 = 231.533 g/mol
According to the balanced equation:
1 mole of Fe reacts with 1 mole of Fe3O4
Therefore, the number of moles of Fe3O4 that can be formed is also 12.34 mol (since it is limited by the available moles of Fe).
Now, we need to find the mass of Fe3O4:
Mass = Number of moles × Molar mass
Mass of Fe3O4 = 12.34 mol × 231.533 g/mol = 2,853.58 g
Hence, the maximum amount of Fe3O4 that can be formed is 2,853.58 g.
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To find the mass of water needed to react completely with Na2O2, we need to consider the stoichiometry of the reaction and the molar ratios.
Given:
Molar mass of Na2O2 = 77.98 g/mol
According to the balanced equation:
2 moles of Na2O2 react with 1 mole of H2O
Therefore, the molar ratio between Na2O2 and H2O is 2:1.
To find the amount of water needed, we need to know the amount of Na2O2.
If you provide the mass or number of moles of Na2O2, I can help you calculate the mass of water needed.
To answer these questions, we need to use stoichiometry and molar ratios.
1. To find the amount of Na3PO4 produced from 3.43 g of H3PO4, we first need to convert grams of H3PO4 to moles by using its molar mass. The molar mass of H3PO4 is 97.994 g/mol.
Number of moles of H3PO4 = mass / molar mass
Number of moles of H3PO4 = 3.43 g / 97.994 g/mol
Next, we use the balanced equation to determine the molar ratio between H3PO4 and Na3PO4. The balanced equation tells us that 1 mole of H3PO4 reacts with 1 mole of Na3PO4.
So, the number of moles of Na3PO4 = number of moles of H3PO4
We can then convert the moles of Na3PO4 to grams by using its molar mass. The molar mass of Na3PO4 is 163.94 g/mol.
Mass of Na3PO4 = number of moles of Na3PO4 * molar mass of Na3PO4
2. To find the maximum amount of Fe3O4 that could be formed, we need to determine the limiting reactant. This can be done by comparing the number of moles of Fe and the number of moles of H2O.
We need to use the balanced equation to determine the molar ratio between the reactants Fe and H2O. From the equation, we can see that 4 moles of Fe react with 3 moles of H2O to produce 1 mole of Fe3O4.
Next, we calculate the number of moles of Fe3O4 that can be formed from each reactant.
From 12.34 mol of Fe, the maximum number of moles of Fe3O4 = 12.34 mol * (1 mol Fe3O4 / 4 mol Fe)
From 8.74 mol of H2O, the maximum number of moles of Fe3O4 = 8.74 mol * (1 mol Fe3O4 / 3 mol H2O)
The limiting reactant is the one that produces the smallest amount of Fe3O4. So, we use the smaller value obtained from the above calculations.
Finally, we can convert the moles of Fe3O4 to grams by using its molar mass.
Mass of Fe3O4 = number of moles of Fe3O4 * molar mass of Fe3O4
3. To find the mass of water needed to react completely with Na2O2, we need to use the balanced equation.
The balanced equation for the reaction is: Na2O2 + 2H2O -> 2NaOH + H2O
From the equation, we can see that 2 moles of water react with 1 mole of Na2O2.
To determine the mass of water needed, we first need to convert the moles of Na2O2 to moles of water by using the molar ratio.
Then, we can convert the moles of water to grams by using the molar mass of water.
Mass of water = number of moles of water * molar mass of water
I hope this helps your friend! If there are any further questions or clarifications needed, feel free to ask.