Homework 56


Question 9.41:

A) Which of the following compounds are soluble?

a. $Pb(NO_{3})_2$

b. $LiC_{2}H_{3}O_{2}$

c. $NaOH$

d. $H_{2}S$

B) Determine the ions that break up from the the soluble compound in a solution.

Solution:

A)

a. soluble: Based on solubility rules, $NO_{3}^{-}$ is always soluble. There are no exceptions. Therefore, this aqueous compound will ionize and break into ions.

b. soluble: Based on solubility rules, any compound containing $Li^{+}$ is soluble and will break into ions. $C_{2}H_{3}O_{2}^{-}$ is also always soluble.

c. soluble: Based on solubility rules, $OH^{-}$ is generally insoluble. However, $Na^{+}$ is an exception to this rule. Since this compound contains $Na^{+}$, it will break into ions.

d. insoluble: Based on the solubility rules, $S^{2-}$ is generally insoluble. $H^{+}$ is not an exception to this rule. Therefore, this compound will form a solid (or a precipitate) and will not break into ions.

B)

a. $Pb^{2+}; 2NO_{3}^{-}$

b. $Li^{+}; C_{2}H_{3}O_{2}^{-}$

c. $Na^{+}; OH^{-}$

d. This compound does not break into ions. Instead, since it is insoluble, it is a precipitate and is a solid.

Question 10.55:

After a complete reaction of 6.50kg of $CH_{4}$, how much heat formed?

$CH_{4(g)} + 2O_{2(g)} \rightarrow CO_{2(g)} + 2H_{2}O_{(g)}$

$\Delta H_{rxn} = -1250 kJ$

Solution:

Step 1: Is the equation balanced? Yes. Also note that there is 1 mol of $CH_{4}$ in this reaction. There are 2 mol of $O_{2}$ in this reaction. There is 1 mol of $CO_{2}$ in this reaction. There are 2 mol of $H_{2}O$ in this reaction.

Step 2: What is given? We know that there is 6.50kg of $CH_{4}$.

We know that $\Delta H_{rxn} = -1250 kJ$

Step 3: What other conversion factors might be useful to solve this problem?

1,000g=1kg

The molar mass of $CH_{4}$=16.05g/mol (Use the periodic table to locate the molar mass for $C$ and $H$. Multiply the molar mass of hydrogen by 4. Add this number, 4.04g/mol, to the molar mass of $C$,12.01g/mol.)

Step 4: Set up dimensional analysis, starting with the value given in the question. (Make sure that your units "match" when completing this step.

$6.50kg CH_{4}\times \frac{1000g CH_{4}}{1kg CH_{4}}\times \frac{1 mol CH_{4}}{16.05g CH_{4}}\times\frac{-1250kJ}{1 mol CH_{4}} = -5.06\times10^{4}kJ$

The negative sign indicates that this is an exothermic reaction. In other words, the system (the reactants) lost energy to the surroundings (the products) in the form of heat.

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