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Chemical Reactions and Stoichiometry Problems

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Please assist with answering the following questions. I'd love to be able to see a step to step process for solving these, that way when I come across problems similar to this in the future and then, I don't get lost again.

1. 2Mn02 + 4KOH + O2 + CL2 ---> 2KMnO4 + 2KCL + 2H2O
For the reaction above, there is 100.0 g of each reactant available. Which reagent is the limiting reagent?

2. The reaction of 7.8 g of benzene (C6H6) with excess nitric acid resulted in 0.90 g of H2O. What is the percentage yield?
Molar mass (g/mol): C6H6=78, HNO3=63, C6H5NO2=123, H2O=18
C6H6 + HNO3 ----> C6H5NO2 + H2O

3. 6H+ + 5H2O + 2MnO4- ----> 5O2 + 2Mn2+ + 8H2O
According to the balanced equation above, how many moles of permanganate ion are required to react completely with 25.0 ml of .100 M hydrogen peroxide?

4. How many grams of water will be formed when 32.0 g of hydrogen is allowed to react with 16.0 g of oxygen according to the following reaction:
2H2 + O2 = 2H2O

5. What mass of Aluminum is produced when 0.5 mole of Al2S3 is completely reduced with excess H2?

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See explanation in the Word document. Please see the attached file.

1. 2Mn02 + 4KOH + O2 + CL2 ---> 2KMnO4 + 2KCL + 2H2O
For the reaction above, there is 100.0g of each reactant available. Which reagent is the limiting reagent?

First, in the balanced reaction, the coefficients mean that for 2 mol of MnO2 to be consumed completely, 4 mol of KOH, 1 mol of O2 and 1 mol of Cl2 are needed.
Based on the definition of limiting reagent, it is the one that is consumed completely.
Then we need to figure out the molar amount for each reactant. To do that, we need to figure out the molar mass for each reactant:
MnO2: 54.9+16.0*2=86.9 g/mol (Note: 54.9 and 16.0 for Mn and O could be found in the periodic table)
KOH: 39.1+16.0+1.0=56.1 g/mol.
O2: 16.0*2=32.0 g/mol
Cl2: 35.5*2=71.0 g/mol
Once we have the molar mass, we use the formula mass/molar mass to find out the molar amount:
MnO2: 100/86.9=1.15 mol
KOH: 100/56.1=1.78 mol
O2: 100/32.0=3.13 mol
Cl2: 100/71.0=1.41 mol
Then we find out the ratio between these reactants:
1.15:1.78:3.13:1.41=0.8:1.3:2.2:1 (Note: I make the Cl2 ratio as 1 to make it comparable to the ratio in ...

Solution Summary

The solution discusses answers to the chemical reactions and stoichiometry problems.

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Stoichiometry Problems and Chemical Equations

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1) Calculate the weight of 50% NaOH solution required to prepare 500 ml of 0.1 M NaOH.

2) Calculate the volume of 19.1 M NaOH (50% weight) solution required to prepare 500 mol of 0.1 m NaOH. (Density of 50% NaOH solution is 1.5 g/mL.)

3) Write the chemical equations for the preparation of an amide from an organic acid (RCO2H).

4) What weight of potassium acid phtalate (KHP) is required for the neutralization of 45.0 mL of 0.095 M NaOH?

5) Calculate the equivalent weight of 4-touluic acid.

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