{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

H03- Mixtures and KMT-solutions

# H03- Mixtures and KMT-solutions - moczulski(ksm935 – H03...

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

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: moczulski (ksm935) – H03: Mixtures and KMT – mccord – (50960) 1 This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. All students from Dr. McCord’s MWF 11am class (50960) must report to the correct room for the exams. There are 2 different rooms and the class is split according to the first letter of your last name. Go to the right room or you WILL receive a zero on the exams. A-L go to UTC 2.112A M-Z go to UTC 2.102A 001 (part 1 of 4) 10.0 points Iron pyrite (FeS 2 ) is the form in which much of the sulfur exists in coal. In the combustion of coal, oxygen reacts with iron pyrite to produce iron(III) oxide and sulfur dioxide, which is a major source of air pollution and a substantial contributor to acid rain. What mass of Fe 2 O 3 is produced from the reaction is 67 L of oxygen at 2 . 79 atm and 152 ◦ C with an excess of iron pyrite? Correct answer: 155 . 621 g. Explanation: P = 2 . 79 atm T = 152 ◦ C + 273 = 425 K R = 0 . 08206 L · atm K · mol V = 67 L MW Fe 2 O 3 = 2(55 . 845 g / mol) + 3(15 . 9994 g / mol) = 159 . 688 g / mol The balanced equation is 4 FeS 2 (s) + 11 O 2 (g)-→ 2 Fe 2 O 3 (s) + 8 SO 2 (g) Applying the ideal gas law to the O 2 , P V = nRT n = P V RT = (2 . 79 atm) (67 L) ( . 08206 L · atm K · mol ) (425 K) = 5 . 35992 mol . From stoichiometry and the molar mass of Fe 2 O 3 , m Fe 2 O 3 = (159 . 688 g / mol Fe 2 O 3 ) × 2 mol Fe 2 O 3 11 mol O 2 (5 . 35992 mol O 2 ) = 155 . 621 g Fe 2 O 3 . 002 (part 2 of 4) 10.0 points If the sulfur dioxide that is generated above is dissolved to form 4 . 1 L of aqueous solu- tion, what is the molar concentration of the resulting sulfurous acid (H 2 SO 3 ) solution? Correct answer: 0 . 950762 M. Explanation: V = 4 . 1 L SO 2 (g) + H 2 O( ℓ )-→ H 2 SO 3 (aq) . From the stoichiometry, n SO 2 = (5 . 35992 mol) parenleftbigg 8 n SO 2 11 n O 2 parenrightbigg = 3 . 89813 mol . 3 . 89813 mol of SO 2 will dissolve in 4 . 1 L of water to form a solution that is 3 . 89813 mol 4 . 1 L = 0 . 950762 M in H 2 SO 4 . 003 (part 3 of 4) 10.0 points What mass of SO 2 is produced in the burning of 1 tonne (1 t = 1000 kg) of high-sulfur coal, if the coal is 6% pyrite by mass? Correct answer: 64 . 0771 kg. Explanation: m coal = 1000 kg m FeS 2 = 6%(1000 kg) = 60 kg = 60000 g MW FeS 2 = 55 . 845 g / mol + 2(32 . 065 g / mol) = 119 . 975 g MW SO 2 = 32 . 065 g / mol + 2(15 . 9994 g / mol) = 64 . 0638 g moczulski (ksm935) – H03: Mixtures and KMT – mccord – (50960) 2 m SO 2 = 1000 kg coal parenleftbigg 60000 g FeS 2 1000 kg coal parenrightbigg × parenleftbigg 1 mol FeS 2 119 . 975 g FeS 2 parenrightbigg × parenleftbigg 8 mol SO 2 4 mol FeS 2 parenrightbiggparenleftbigg 64 . 0638 g SO 2 1 mol SO 2 parenrightbigg = 64077 . 1 g = 64 . 0771 kg SO 2 ....
View Full Document

{[ snackBarMessage ]}

### Page1 / 7

H03- Mixtures and KMT-solutions - moczulski(ksm935 – H03...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online