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Chapter 3: Problem 45

The formula for rust can be represented by \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) How many moles of Fe are present in \(24.6 \mathrm{~g}\) of the compound?

Short Answer

Expert verified
So, there are 0.308 moles of Fe in 24.6 g of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\).

Step by step solution

01

Determine the Molar Mass of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\)

The molar mass of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) can be calculated by adding up the molar masses of all atoms in the molecule. The molar mass of iron (Fe) is 55.85 g/mol and that of oxygen (O) is 16.00 g/mol. Hence, the molar mass of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) = \(2 * 55.85 \mathrm{g/mol} + 3 * 16.00 \mathrm{g/mol} = 159.70 \mathrm{g/mol}\).
02

Convert the Given Mass to Moles

The number of moles can be found using the relation moles = mass / molar mass. Plugging in the given mass of the compound \(24.6 \mathrm{g}\) and the calculated molar mass, the number of moles of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) = \(24.6 \mathrm{g} / 159.70 \mathrm{g/mol} = 0.154 \mathrm{moles}\).
03

Determine the Moles of Iron (Fe)

One formula unit of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) has two moles of iron (Fe). Therefore, the number of moles of Fe in the given compound = \(2 * 0.154 \mathrm{moles} = 0.308 \mathrm{moles}\).

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Most popular questions from this chapter

Monosodium glutamate (MSG), a food-flavor enhancer, has been blamed for "Chinese restaurant syndrome," the symptoms of which are headaches and chest pains. MSG has the following composition by mass: 35.51 percent \(\mathrm{C}, 4.77\) percent \(\mathrm{H}, 37.85\) percent \(\mathrm{O}, 8.29\) percent \(\mathrm{N},\) and 13.60 percent \(\mathrm{Na}\). What is its molecular formula if its molar mass is about \(169 \mathrm{~g} ?\)

If we know the empirical formula of a compound, what additional information do we need to determine its molecular formula?

Heating \(2.40 \mathrm{~g}\) of the oxide of metal \(\mathrm{X}\) (molar mass of \(X=55.9 \mathrm{~g} / \mathrm{mol}\) ) in carbon monoxide \((\mathrm{CO})\) yields the pure metal and carbon dioxide. The mass of the metal product is \(1.68 \mathrm{~g}\). From the data given, show that the simplest formula of the oxide is \(\mathrm{X}_{2} \mathrm{O}_{3}\) and write a balanced equation for the reaction.

Nitroglycerin \(\left(\mathrm{C}_{3} \mathrm{H}_{5} \mathrm{~N}_{3} \mathrm{O}_{9}\right)\) is a powerful explosive. Its decomposition can be represented by $$ 4 \mathrm{C}_{3} \mathrm{H}_{5} \mathrm{~N}_{3} \mathrm{O}_{9} \longrightarrow 6 \mathrm{~N}_{2}+12 \mathrm{CO}_{2}+10 \mathrm{H}_{2} \mathrm{O}+\mathrm{O}_{2} $$ This reaction generates a large amount of heat and many gaseous products. It is the sudden formation of these gases, together with their rapid expansion, that produces the explosion. (a) What is the maximum amount of \(\mathrm{O}_{2}\) in grams that can be obtained from \(2.00 \times 10^{2} \mathrm{~g}\) of nitroglycerin? (b) Calculate the percent yield in this reaction if the amount of \(\mathrm{O}_{2}\) generated is found to be \(6.55 \mathrm{~g}\).

Explain clearly what is meant by the statement "The atomic mass of gold is 197.0 amu."
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