Question

10. Iron (III) oxide is used as a source material for iron in steel manufacturing. The density ofiron (III) oxide is 5.242 g/cm'.a. If a plant needed to transport 88,915 kg of iron(III) oxide every day for operations, whatvolume of iron(III) oxide is transported (in Liters)?10a.b. If a standard, tandem axle dump truck can haul 15 tons of ore and the mine is one dayaway from the manufacturing site, how many dump trucks are needed each day?ti10b.c. Each dump truck must drive 285 miles from the source to the manufacturing site. If eachtruck gets 7.8 miles per gallon of gas and the price of gas is $3.95 per gallon, what is thefuel cost, per day, for the manufacturing plant?10c.

Answer 1

The volume of iron (III) oxide is 16962.04 Liters

Explanation:

Given information

The density of iron (III) oxide = 5.242 g/cm^3

The mass of iron (III) oxide = 88,915 kg

Let x represents the volume in liters of iron (III) oxide

The first step is to convert the mass from kilograms to grams for unit consistency since the density is already measured in g/cm^3

Recall that, 1 kg is equivalent to 1000 g according to the standard international unit

let x represents the mass to be converted to grams

`\begin{gathered} 1kg\text{ }\rightarrow\text{ 1000 grams} \\ 88915kg\text{ }\rightarrow\text{ xgrams} \\ \text{Cross multiply} \\ 1\cdot\text{ x = 1000 }\cdot\text{ 88915} \\ x\text{ = 88915000 grams} \end{gathered}`

From the calculation above, you will see that the mass of iron (III) oxide in grams is 88915000 grams.

The next step is to find the volume of iron (III) oxide using the below formula

`\text{ Density = }\frac{mass}{\text{volume}}`
`\begin{gathered} \text{Isolate volume} \\ \text{mass = density x volume} \\ \text{volume = }\frac{mass}{\text{density}} \end{gathered}`
`\begin{gathered} \text{Volume = }(88915000)/(5.242) \\ \text{Volume = 16962037.39}cm^3 \end{gathered}`

The next step is to convert the cubic centimeters to liters

According to the standard international unit, 1 cm^3 is equivalent to 0.001 L

`\begin{gathered} 1cm^3\text{ }\rightarrow\text{ 0.001L} \\ 16962307.39cm^3\text{ }\rightarrow\text{ xL} \\ \text{Cross multiply} \\ 1\cdot\text{ x = 0.001 }\cdot\text{ 1}6962307.39 \\ x\text{ = 16962.04 Liters} \end{gathered}`

Therefore, the volume of iron (III) oxide is 16962.04 Liters