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Chapter 3: Problem 13

Calculate each of the following quantities: (a) Amount (mol) of Mn atoms in \(62.0 \mathrm{mg}\) of Mn (b) Amount (mol) for \(1.36 \times 10^{22}\) atoms of \(\mathrm{Cu}\) (c) Mass (g) of \(8.05 \times 10^{24} \mathrm{Li}\) atoms

Short Answer

Expert verified
a) 0.00113 mol, b) 0.0226 mol, c) 92.81 g

Step by step solution

01

Convert Mass of Mn to Moles

To find the amount of Mn atoms in 62.0 mg of Mn, first convert the mass from mg to grams. \[ 62.0 \text{ mg} = 0.0620 \text{ g} \]Next, use the molar mass of Mn, which is approximately 54.94 g/mol, to convert grams to moles. \[ \text{moles of Mn} = \frac{0.0620 \text{ g}}{54.94 \text{ g/mol}} \]Calculate the result. \[ \text{moles of Mn} ≈ 0.00113 \text{ mol} \]
02

Convert Number of Cu Atoms to Moles

To find the amount of moles for 1.36 × 10^{22} atoms of Cu, use Avogadro's number, which is \(6.022 \times 10^{23}\) atoms/mol.\[ \text{moles of Cu} = \frac{1.36 \times 10^{22} \text{ atoms}}{6.022 \times 10^{23} \text{ atoms/mol}} \]Calculate the result. \[ \text{moles of Cu} ≈ 0.0226 \text{ mol} \]
03

Convert Number of Li Atoms to Mass

To find the mass of 8.05 × 10^{24} Li atoms, first convert the number of atoms to moles using Avogadro's number.\[ \text{moles of Li} = \frac{8.05 \times 10^{24} \text{ atoms}}{6.022 \times 10^{23} \text{ atoms/mol}} \]Calculate the moles. \[ \text{moles of Li} ≈ 13.37 \text{ mol} \]Next, use the molar mass of Li, which is approximately 6.94 g/mol, to convert moles to grams.\[ \text{mass of Li} = 13.37 \text{ mol} \times 6.94 \text{ g/mol} \]Calculate the result. \[ \text{mass of Li} ≈ 92.81 \text{ g} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

molar mass
Molar mass is a fundamental concept in chemistry. It allows us to convert between the mass of a substance and the amount (in moles) of that substance. The molar mass of an element is the mass of one mole of its atoms, typically measured in grams per mole (g/mol). It is equivalent to the atomic or molecular weight of the element or compound, scaled up to Avogadro's number of particles.

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Most popular questions from this chapter

3.58 Write balanced equations for each of the following by inserting the correct coefficients in the blanks: (a) \(-\mathrm{Cu}(s)+\underline{\mathrm{S}}_{8}(s) \longrightarrow-\mathrm{Cu}_{2} \mathrm{~S}(s)\) (b) \(-\mathrm{P}_{4} \mathrm{O}_{10}(s)+\underline{\mathrm{H}}_{2} \mathrm{O}(l) \longrightarrow-\mathrm{H}_{3} \mathrm{PO}_{4}(l)\) (c) \(-\mathrm{B}_{2} \mathrm{O}_{3}(s)+-\mathrm{NaOH}(a q) \longrightarrow-\mathrm{Na}_{3} \mathrm{BO}_{3}(a q)+-\mathrm{H}_{2} \mathrm{O}(l)\) (d) \(-\mathrm{CH}_{3} \mathrm{NH}_{2}(g)+\underline{\mathrm{O}_{2}}(g) \longrightarrow\) \(-\mathrm{CO}_{2}(g)+-\mathrm{H}_{2} \mathrm{O}(g)+-\mathrm{N}_{2}(g)\)

Metal hydrides react with water to form hydrogen gas and the metal hydroxide. For example, $$ \operatorname{SrH}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \operatorname{Sr}(\mathrm{OH})_{2}(s)+2 \mathrm{H}_{2}(g) $$ You wish to calculate the mass (g) of hydrogen gas that can be prepared from \(5.70 \mathrm{~g}\) of \(\mathrm{SrH}_{2}\) and \(4.75 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\) (a) What amount (mol) of \(\mathrm{H}_{2}\) can be produced from the given mass of \(\mathrm{SrH}_{2} ?\) (b) What amount (mol) of \(\mathrm{H}_{2}\) can be produced from the given mass of \(\mathrm{H}_{2} \mathrm{O} ?\) (c) Which is the limiting reactant? (d) How many grams of \(\mathrm{H}_{2}\) can be produced?

Chlorine gas can be made in the laboratory by the reaction of hydrochloric acid and manganese(IV) oxide: $$ 4 \mathrm{HCl}(a q)+\mathrm{MnO}_{2}(s) \longrightarrow \mathrm{MnCl}_{2}(a q)+2 \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{Cl}_{2}(g) $$ When 1.82 mol of \(\mathrm{HCl}\) reacts with excess \(\mathrm{MnO}_{2}\), how many (a) moles of \(\mathrm{Cl}_{2}\) and (b) grams of \(\mathrm{Cl}_{2}\) form?

Convert the following descriptions into balanced equations: (a) In a gaseous reaction, hydrogen sulfide burns in oxygen to form sulfur dioxide and water vapor. (b) When crystalline potassium chlorate is heated to just above its melting point, it reacts to form two different crystalline compounds, potassium chloride and potassium perchlorate. (c) When hydrogen gas is passed over powdered iron(III) oxide, iron metal and water vapor form. (d) The combustion of gaseous ethane in air forms carbon dioxide and water vapor. (e) Iron(II) chloride is converted to iron(III) fluoride by treatment with chlorine trifluoride gas. Chlorine gas is also formed.

A 0.652 -g sample of a pure strontium halide reacts with excess sulfuric acid. The solid strontium sulfate formed is separated, dried, and found to weigh \(0.755 \mathrm{~g}\). What is the formula of the original halide?
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