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The rock in a particular iron ore deposit contains 78% Fe2O3 by mass. How many kilograms
Chapter 6, Problem 111P(choose chapter or problem)
The rock in a particular iron ore deposit contains 78% \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) by mass. How many kilograms of the rock must a mining company process to obtain \(1.0 \times 10^{3}\) kg of iron?
Equation Transcription:
Text Transcription:
Fe_2 O_3
1.0 times 10^3
Questions & Answers
QUESTION:
The rock in a particular iron ore deposit contains 78% \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) by mass. How many kilograms of the rock must a mining company process to obtain \(1.0 \times 10^{3}\) kg of iron?
Equation Transcription:
Text Transcription:
Fe_2 O_3
1.0 times 10^3
ANSWER:
Solution 111P
Here, we are going to calculate the mass of rock required by the mining company to obtain 1.0 x 103 kg of iron.
Step 1:
Molar mass of Fe2O3 = (2 x atomic mass of Fe) + (3 x atomic mass of O)
= (2 x 55.845) + (3 x 15.9994)
= 111.69 + 47.9982
= 159.6882 g/mol