# Help with these 4 questions?!

## 1) For the reaction of aluminum with conc. sulfuric acid, $\left(a\right)$ balance the reaction, $\left(b\right)$ determine the volume in $\text{L}$ of ${\text{H}}_{2}$ produced if $\text{10.00 g}$ of $\text{Al}$ reacts with excess acid, and $\left(c\right)$ determine the mass in $\text{g}$ of $\text{Al} \left(s\right)$ that $\text{25.00 mL}$ of ${\text{1.244 M H"_2"SO}}_{4}$ should react with. 2) For the reaction of phosphoricacid with magnesium hydroxide, $\left(a\right)$ write the balanced reaction, and $\left(b\right)$ find the mass in $\text{g}$ of magnesium hydroxide that would react with $\text{25.00 mL}$ of $\text{0.4376 M}$ of phosphoric acid. 3) How much water do you predict would have to be added to dilute a $\text{3.0 M}$ aqueous solution of $\text{HCl}$ to yield $\text{100.0 mL}$ of a $\text{0.135 M}$ solution? Assume additivity of solution volumes. 4) For the reaction of potassium sulfide and cobalt (II) nitrate, $\left(a\right)$ write the balanced chemical reaction, $\left(b\right)$ determine the limiting reagent and find the mass in $\text{g}$ of cobalt (II) sulfide formed from the reaction of $\text{175 mL}$ of $\text{0.225 M}$ potassium sulfide with $\text{250 mL}$ of $\text{0.180 M}$ cobalt (II) nitrate.

May 29, 2017

Well, here's $\left(1\right)$. The concepts are very similar in each of these problems, except for $\left(3\right)$, which is a dilution.

1)

a) Generally when you start with a pure metal and get a gas, it is a redox reaction. It is also a single-replacement reaction, of the form

$A + B C \to B + A C$.

• Aluminum implies $\text{Al} \left(s\right)$, the metal element.
• Sulfuric acid is a sulfur-based acid with the "sulfate" polyatomic anion, ${\text{SO}}_{4}^{2 -}$, so this is a diprotic acid, ${\text{H"_2"SO}}_{4}$.
• Hydrogen naturally exists as the diatomic gas, ${\text{H}}_{2} \left(g\right)$.

So, we write the unbalanced reaction as:

"Al"(s) + "H"_2"SO"_4(l) -> "H"_2(g) + ???

In fact, you can tell that if an acid (which is itself dissolved in some water) dissolves a metal, the metal has become a cation (which dissolves well in water!), i.e. it got oxidized by the acid.

Hence (and we include the counterion ${\text{SO}}_{4}^{2 -}$ to ${\text{Al}}^{3 +}$), we get "Al"_2("SO"_4)_3 as the ???:

"Al"(s) + "H"_2"SO"_4(l) -> "H"_2(g) + "Al"_2("SO"_4)_3(aq)

Now, balancing is fairly straightforward; treat the sulfate as an entire unit and add a coefficient in front of the sulfuric acid in order to balance the sulfate itself.

Then $\text{Al}$ and ${\text{H}}_{2}$ will nicely balance as they are pure elements on the reactants and products side, respectively (it is easy in the sense that you won't disturb the mass balance of other elements).

color(blue)(2"Al"(s) + 3"H"_2"SO"_4(l) -> 3"H"_2(g) + "Al"_2("SO"_4)_3(aq))

b) From $\text{10.00 g Al}$ at STP, ${0}^{\circ} \text{C}$ and $\text{1 atm}$, we have:

10.00 cancel"g Al" xx cancel"1 mol Al"/(26.981 cancel"g Al") xx "3 mol H"_2/(2 cancel"mol Al")

$= {\text{0.5559 mol H}}_{2}$

This means that via the ideal gas law, $P V = n R T$, you would get from $\text{273.15 K}$ and $\text{1 atm}$ for STP, and a universal gas constant $R = \text{0.082057 L"cdot"atm/mol"cdot"K}$:

$\textcolor{b l u e}{{V}_{{H}_{2} \left(g\right)}} = \frac{n R T}{P}$

$= \left(\text{0.5559 mol H"_2cdot"0.082057 L"cdot"atm/mol"cdot"K"cdot"273.15 K")/("1 atm}\right)$

$= \textcolor{b l u e}{{\text{12.46 L H}}_{2}}$

c) Similarly, we find the mols of sulfuric acid available using its volume and concentration.

$\left(\text{1.244 mol H"_2"SO"_4)/cancel"L" xx 25.00 cancel"mL" xx cancel"1 L"/(1000 cancel"mL}\right)$

$= {\text{0.0311 mols H"_2"SO}}_{4}$

This means that you needed:

0.0311 cancel("mols H"_2"SO"_4) xx "2 mols Al"/(3 cancel("mols H"_2"SO"_4))

$=$ $\text{0.02073 mols Al}$

Using the molar mass of $\text{Al}$ above, show that you would get $\textcolor{b l u e}{\text{0.5594 g Al}}$.

May 29, 2017

Problem (4): $\text{3.6 g CoS}$

#### Explanation:

Potassium sulfide and cobalt(II) nitrate are soluble ionic compounds, which means that they dissociate completely when dissolved in water.

The cobalt(II) cations and the sulfide anions will combine to form cobalt sulfide, an insoluble ionic compound that will precipitate out of solution.

The second product of the reaction will be aqueous potassium nitrate, a soluble ionic compound that will exist as ions in the resulting solution.

So, you can say that the balanced chemical equation that describes this double replacement reaction looks like this

${\text{K"_ 2"S"_ ((aq)) + "Co"("NO"_ 3)_ (2(aq)) -> "CoS"_ ((s)) darr + 2"KNO}}_{3 \left(a q\right)}$

Now, use the molarities and the volumes of the two solutions to determine how many moles of potassium sulfide and cobalt(II) nitrate are being mixed

175 color(red)(cancel(color(black)("mL solution"))) * ("0.225 moles K"_2"S")/(10^3color(red)(cancel(color(black)("mL solution")))) = "0.03938 moles K"_2"S"

250 color(red)(cancel(color(black)("mL solution"))) * ("0.180 mole Co"("NO"_3)_2)/(10^3color(red)(cancel(color(black)("mL solution")))) = "0.04500 moles Co"("NO"_3)_2

According to the balanced chemical equation, the reaction consumes potassium sulfide and cobalt(II) nitrate in a $1 : 1$ mole ratio.

As you can see, you have $0.04500$ moles of cobalt(II) nitrate and only $0.03938$ moles of potassium sulfide. This implies that potassium sulfide will be completely consumed before all the moles of cobalt(II) nitrate will get the chance to react $\to$ potassium sulfide will act as a limiting reagent.

Therefore, you can say that the reaction will consume $0.03938$ moles of potassium sulfide and of cobalt(II) nitrate, and produce $0.03938$ moles of cobalt sulfide.

To convert this to grams, use the compound's molar mass

$0.03938 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles CoS"))) * "90.998 g"/(1color(red)(cancel(color(black)("mole CoS")))) = color(darkgreen)(ul(color(black)("3.6 g}}}}$

The answer is rounded to two sig figs, the number of sig figs you have for the volume of the cobalt(II) nitrate solution.

May 29, 2017

#### Explanation:

Phosphoric acid is tribasic and has the formula $\textsf{{H}_{3} P {O}_{4}}$.

Magnesium hydroxide is sf(Mg(OH)_2.

The essential reaction occurring is:

$\textsf{{H}^{+} + O {H}^{-} \rightarrow {H}_{2} O}$

To ensure that these balance we will need 2 moles of phosphoric acid for every 3 moles of magnesium hydroxide.

The balanced equation is:

$\textsf{\textcolor{red}{2} {H}_{3} P {O}_{4 \left(a q\right)} + \textcolor{red}{3} M g {\left(O H\right)}_{2 \left(s\right)} \rightarrow M {g}_{3} {\left(P {O}_{4}\right)}_{2 \left(a q\right)} + 6 {H}_{2} {O}_{\left(l\right)}}$

We know that concentration = moles/volume

$\textsf{c = \frac{n}{v}}$

From the titration result we can get the no. moles of phosphoric acid:

$\textsf{{n}_{{H}_{3} P {O}_{4}} = c \times v = 0.4376 \times \frac{25.00}{1000} = 0.01094}$

From the equation we can say that the no. moles of magnesium hydroxide will 3/2 x this amount.

$\therefore$$\textsf{{n}_{M g {\left(O H\right)}_{2}} = \frac{3}{2} \times 0.01094 = 0.01641}$

The $\textsf{{M}_{r}}$ of magnesium hydroxide is 58.32.

This means 1 mole weighs 58.32 g.

So the mass of magnesium hydroxide is given by:

$\textsf{m = n \times {M}_{r} = 0.01641 \times 58.32 = 0.1365 \textcolor{w h i t e}{x} g}$

May 29, 2017

Question $3$:

$95.5 \text{mL H"_2"O}$ needed

#### Explanation:

We're asked to find the amount of water needed to dilute a $3.0 M$ solution of $\text{HCl}$ to produce $100.0 \text{mL}$ of a $0.135 M$ $\text{HCl}$ solution.

To solve this, we can use the dilution formula,

${M}_{\text{conc"V_"conc" = M_"dil"V_"dil}}$

We need to find the volume of the concentrated $\text{HCl}$ solution, ${V}_{\text{conc}}$, so let's rearrange the equation to solve for ${V}_{\text{conc}}$:

V_"conc" = (M_"dil"V_"dil")/(M_"conc")

Now, let's plug in our known variables and find the volume of the concentrated solution:

${V}_{\text{conc" = ((0.135cancel("mol"/"L"))(100.0"mL"))/(3.0cancel("mol"/"L")) = color(red)(4.5"mL}}$

This represents the volume of the concentrated $3.0 M$ $\text{HCl}$ solution; we were asked to find how much water needed to be added. Finding this is simple enough, we just subtract the initial volume ($4.5 \text{mL}$) from the final volume ($100.0 \text{mL}$):

$\text{H"_2"O needed" = 100.0"mL" - color(red)(4.5"mL") = color(blue)(95.5"mL H"_2"O}$

Thus, you need to add $95.5 \text{mL}$ of water to dilute the $\text{HCl}$ solution to $100.0 \text{mL}$ and $0.135 M$.