{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

H02- Stoichiometry-solutions

# H02- Stoichiometry-solutions - truitt(wtt255 H02...

This preview shows pages 1–3. Sign up to view the full content.

truitt (wtt255) – H02: Stoichiometry – Mccord – (50970) 1 This print-out should have 29 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. This assignment comes from Chapters 2 and 3 in your book. Nomenclature (the nam- ing of compounds, etc...) is in Chapter 2. Chapter 3 is on stoichiometry. You need to know how to calculate all the types of prob- lems on this homework. ALEKS does do this well. READ Chapter 3 and LOOK at the ex- amples. This is important although I will not formally cover it in class. You are expected to know this already. 001 10.0 points Which compound has the wrong chemical for- mula? 1. Ba 3 (PO 4 ) 2 2. (NH 4 ) 2 SO 4 3. Mg(OH) 2 4. CaOH correct Explanation: The calcium ion is Ca 2+ ; the hydroxide ion is OH - . Two OH - are needed to balance the charge of each Ca 2+ , so the formula is Ca(OH) 2 . The magnesium ion is Mg 2+ ; the hydroxide ion is OH - . Two OH - are needed to balance the charge of each Mg 2+ , so the formula is Mg(OH) 2 . The barium ion is Ba 2+ ; the phosphate ion is PO 3 - 4 . Two PO 3 - 4 are needed to balance the charge of every three Ba 2+ . (This gives a total anion charge of - 6 and a total cation charge of +6.) The formula is Ba 3 (PO 4 ) 2 . The ammonium ion is NH + 4 ; the sulfate ion is SO 2 - 4 . Two NH + 4 are needed to balance the charge of each SO 2 - 4 , so the formula is (NH 4 ) 2 SO 4 . 002 10.0 points Cobalt(III) oxide would be the label for 1. CoO 3 . 2. Co 3 O 2 . 3. Co 3 O. 4. Co 2 O 3 . correct 5. CoO. Explanation: The cobalt(III) ion is Co 3+ ; the oxide ion is O 2 - . Three O 2 - are needed to balance the charge of every two Co 3+ . (This gives a total cation charge of +6 and a total anion charge of - 6.) The formula is Co 2 O 3 . 003 10.0 points Give the formula for aluminum hydroxide. 1. Al(OH) 3 correct 2. AlOH 3. AlOH 3 4. AlH 3 Explanation: The aluminum ion is Al 3+ ; the hydroxide ion is OH - . Three OH - are needed to balance the charge of each Al 3+ , so the formula is Al(OH) 3 . 004 10.0 points Choose the formula for the compound potas- sium dichromate. 1. KCr 2 O 7 2. K 2 CrO 4 3. KCrO 4 4. K 2 Cr 3 O 4 5. K 2 Cr 3 6. K 2 Cr 2 O 7 correct

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
truitt (wtt255) – H02: Stoichiometry – Mccord – (50970) 2 7. K 2 Cr 8. KCrO 3 Explanation: 005 10.0 points How many atoms are in 0 . 0468 mol K? Correct answer: 2 . 8183 × 10 22 atoms. Explanation: n K = 0 . 0468 mol n K = ? atoms 0 . 0468 mol K × 6 . 022 × 10 23 atoms K mol K = 2 . 8183 × 10 22 atoms K 006 10.0 points Which one has the greatest number of atoms? 1. 3.05 moles of water 2. 3.05 moles of argon 3. 3.05 moles of helium 4. 3.05 moles of CH 4 correct 5. All have the same number of atoms Explanation: For 3.05 moles of water: ? atoms = 3 . 05 mol H 2 O × 6 . 02 × 10 23 molec 1 mol × 3 atoms 1 molecule = 5 . 51 × 10 24 atoms For 3.05 moles of CH 4 : ? atoms = 3 . 05 mol CH 4 × 6 . 02 × 10 23 molec 1 mol × 5 atoms 1 molecule = 9 . 18 × 10 24 atoms For 3.05 moles of helium: ? atoms = 3 . 05 mol He × 6 . 02 × 10 23 atoms 1 mol = 1 . 84 × 10 24 atoms For 3.5 moles of argon: ? atoms = 3 . 05 mol Ar × 6 . 02 × 10 23 atoms 1 mol = 1 . 84 × 10 24 atoms 007 10.0 points Each mole of Al(NO 3 ) 3 contains how many moles of oxygen atoms?
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern