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# Stoichiometry Problems and Chemical Equations

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I am a bit stumped with the following word problems, and am having trouble setting up the appropriate equations. If anyone can guide me in the right direction, or suggest ways on how to set up the right equations, I would greatly appreciate it:

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.

6) It was found that 19.60 mL of NaOH solution were required to neutralize 10.00 mL of HCL. If the concentration of the base solution was 0.1051 M, caluclate the concentration of the acid solution.

https://brainmass.com/chemistry/composition-stoichiometry/stoichiometry-problems-chemical-equations-436428

#### Solution Preview

Stoichiometry Problems
1) Calculate the weight of 50% NaOH solution required to prepare 500 ml of 0.1 M NaOH.
SOLUTION:
Solving for mole NaOH in 0.1 M solution:
Molarity = M = mole solute / liter solution
0.1 mol/L = mole NaOH / 0.5 L
mole NaOH = 0.05
Converting to gram:
0.05 mole NaOH x (40 g/mole NaOH) = 2 g NaOH
Solving for weight of 50% NaOH solution required to prepare 500 ml of 0.1 M NaOH:
% by weight = (gram solute / gram solution) x 100
50% = (2 g NaOH / weight of NaOH solution) x 100
0.5 = 2 g NaOH / weight of NaOH solution
weight of NaOH solution = 4 g

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

#### Solution Summary

The expert writes chemical equations for preparation of an amide from an organic acid.

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