# Question 59eb8

Nov 9, 2017

See below.

#### Explanation:

The reaction is $F {e}_{2} {O}_{3} + 3 C O \to 2 F e + 3 C {O}_{2}$.
And ,the molar masses(g/mol) for the substances are:
$F e : 55.85$
$F {e}_{2} {O}_{3} : 159.69$
$C O : 28.01$
$C {O}_{2} : 44.01$

Gas constant is
$R = 8.314 \left(J {K}^{-} 1 m o {l}^{-} 1\right)$
$= 0.08206 \left(L a t m {K}^{-} 1 m o {l}^{-} 1\right)$

Temparature is 0℃=273.15K.

Now we're ready to use the ideal gas law: $P V = n R T$.
The volume of gas is $V = \frac{n R T}{P}$.

[Upper row]
$1000$ kg of $F {e}_{2} {O}_{3}$ is
$\frac{1000 \cdot {10}^{3}}{159.69} = 6.262 \cdot {10}^{3}$ mol.

Thus, $6.262 \cdot {10}^{3} \cdot 2 = 1.252 \cdot {10}^{4}$ mol of $F e$ is obtained and its mass is $1.252 \cdot {10}^{4} \cdot 55.85 = 6.99 \cdot {10}^{5}$ (g) $= 699$ (kg). This is right.

You need $6.262 \cdot {10}^{3} \cdot 3 = 1.879 \cdot {10}^{4}$ mol of $C O$ and its mass is $1.879 \cdot {10}^{4} \cdot 28.01 = 5.26 \cdot {10}^{5}$ (g) $= 526$ (kg).

Obtained $C {O}_{2}$ is also $1.879 \cdot {10}^{4}$ mol and its volume is
$V = \frac{1.879 \cdot {10}^{4} \cdot 0.08206 \cdot 273.15}{1} = 4.212 \cdot {10}^{5} \left(L\right) = 421.2 \left({m}^{3}\right)$.
$426.3 \left({m}^{3}\right)$ in the table is a bit too large.

[Lower row]
$104.9$ kg of $F e$ is $\frac{104.9 \cdot {10}^{3}}{55.85} = 1.878 \cdot {10}^{3}$ mol and to obtain this, you need $\frac{1.878 \cdot {10}^{3}}{2} = 9.390 \cdot {10}^{2}$ mol of $F {e}_{2} {O}_{3}$. The mass of $F {e}_{2} {O}_{3}$ is $9.390 \cdot {10}^{2} \cdot 159.69 = 1.500 \cdot {10}^{5}$ (g) =$150.0$ (kg).

The mass of $C O$ is $228.9 - 150.0$ = $78.9$ (kg) and its amount of substance is $\frac{78.9 \cdot {10}^{3}}{28.01} = 2.817 \cdot {10}^{3}$ (mol). This is three times of that of $F e$($9.39 \cdot {10}^{2}$ mol) and matches the chemical equation.

Obtained $C {O}_{2}$ is also $2.817 \cdot {10}^{3}$ mol. Its volume under 25℃,850# mmHg is
$V = \frac{2.817 \cdot {10}^{3} \cdot 0.08206 \cdot \left(273.15 + 25\right)}{\frac{850}{760}}$
$= 6.162 \cdot {10}^{4}$ (L).

Nov 9, 2017

There are no "incorrect" values unless you require a "balanced" outcome, which rarely happens in practice.

#### Explanation:

The "filling in the blanks" can be done with the given data and the balanced chemical equation.
$F {e}_{2} {O}_{3} + 3 C O \to 2 F e + 3 C {O}_{2}$

Converting the mass to moles for iron and the volume to moles for the $C {O}_{2}$ is needed in the first line.

Iron Oxide $\frac{1000}{159.7} = 6.26$ kg-mol (6260 g-mol if you prefer) original

Iron $\frac{699}{55.85} = 12.16$ kg-mol product

The balanced reaction dictates that the $C O$ required is
$\frac{3}{2} \times 12.16 = 18.24$ kg-mol $C O$.

This should be the same for a balanced reaction as
$3 \times 6.26 = 18.78$ Kg-mol for complete reaction with the iron oxide.

The moles of $C {O}_{2} = C O$ in the equation, so we should have 18.24 kg-mol of $C {O}_{2}$ product as well (from the stated Fe product amount). From the ideal gas laws we obtain:
V = nRT/P ; $V = \frac{18240 \times 0.0802 \times 273}{1} = 399 \times {10}^{3} L = 399 {m}^{3}$

You can "mix and match" the other values to decide which ones are "incorrect" OR if you just need to fill the blank with a different number.