# Stoichiometry Problems, Limiting Reagent, Acid/Base, Balancing Chemical Equations

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1. A solution is prepared by dissolving 10.8g ammonium sulfate in enough water to make 100.0mL of stock solution. A 10.00mL sample of this stock solution is added to 50.00mL of water. Calculate the concentration of ammonium ions and sulfate ions in the final solution.

2. One of the most commonly used white pigments in paint is a compound of titanium and oxygen that contains 59.9% Ti by mass. Determine the empirical formula of this compound.

3. A compound that contains only nitrogen and oxygen is 30.4% N by mass; the molar mass of the compound is 92g/mol. What is the empirical formula of the compound? What is the molecular formula of the compound?

4. Mercury and bromide will react with each other to produce mercury (II) bromide:

Hg(l) + Br2(l)  HgBr2(s)

a) What mass of HgBr2 can be produced form the reaction of 10.0g Hg and 9.00g Br2? What mass of which reagent is left unreacted?

b) What mass of HgBr2 can be produced form the reaction of 5.00mL mercury (density = 13.6g/mL) and 4.00mL bromine (density = 3.10g/mL)?

5. Consider the following unbalanced reaction:

Ca3(PO4)2 + H2SO4  CaSO4 + H3PO4

What masses of calcium sulfate and phosphoric acid can be produced from the reaction of 1.0kg calcium phosphate with 1.0kg concentrated sulfuric acid (98% H2SO4 by mass)?

6. What volume of each of the following acids will react completely with 50.00mL of 0.200M NaOH?

a) 0.100 M HCl

b) 0.150 M HNO3

c) 0.200 M HC2H3O2 (1 acidic hydrogen)

Thanks in advance! See attached for full problem.

© BrainMass Inc. brainmass.com September 24, 2018, 2:00 pm ad1c9bdddf - https://brainmass.com/chemistry/acids-and-bases/stoichiometry-problems-limiting-reagent-acid-base-balancing-chemical-equations-109583#### Solution Preview

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1. A solution is prepared by dissolving 10.8g ammonium sulfate in enough water to make 100.0mL of stock solution. A 10.00mL sample of this stock solution is added to 50.00mL of water. Calculate the concentration of ammonium ions and sulfate ions in the final solution.

First, we need to know the molecular formula for ammonium sulfate. It is (NH4)2SO4. Next, we calculate the molar mass. This is determined by adding up the mass of each of the atoms. Here is how we do it:

N x 2 14.0 x 2 = 28.0

H x 8 1.008 x 8 = 8.064

S 32.066

O x 4 16.0 x 4 = 64.0

Total 132.13

Therefore, the molar mass is 132.13 g/mol.

Now, we find out how many moles of ammonium sulfate we added to 100.0 mL of water.

(1 mol/132.13 g)(10.8 g) = 0.0817 mol

Next, we determine the concentration of the original stock solution. To do this, remember that 100.0 mL = 0.1000 L.

0.0817 mol/0.1000 L = 0.817 M

Finally, we take 10.00 mL of this solution and add it to 50.00 mL of water. What's the concentration of this final solution?

First, we determine the total number of moles in 10.00 mL.

(0.817 mol/L)(0.01000 L) = 0.00817 mol

And finally, we put this 0.00817 mol into a total volume of 60.00 ml (the original 10.00 mL plus the 50.00 mL of water).

Therefore, the final concentration is:

0.00817 mol/0.0600 L = 0.136 M

2. One of the most commonly used white pigments in paint is a compound of titanium and oxygen that contains 59.9% Ti by mass. Determine the empirical formula of this compound.

The compound only has titanium and oxygen. 59.9% is titanium. Therefore, 40.1 % is oxygen. Let us assume that we have 100 grams of the substance. Therefore, in 100 grams, we have 59.9 g of Ti and 40.1 g of O. Sound okay so far?

If we have 59.9 grams, how many moles is that? Therefore, we need to know the molar mass of titanium. Look it up in the periodic table. It is 47.88.

(1 mol/47.88 g)(59.9 g) = 1.25 mol

That's a nice number.

Next, we do the same for oxygen.

(1 mol/16.0 g)(40.1 g) = 2.50 mol

That's a nice number.

Therefore, we know that the empirical formula is Ti1.25O2.50. But this is kind of lame looking, isn't it? Therefore, we need whole numbers instead. To do that, we divide each of these numbers by the lowest number. In this case, we divide each by 1.25 like ...

#### Solution Summary

1700 words explain the calculations needed to answer various questions on these acid and base topics.