1.Calculate the mass in grams of 3.37x10^25 molecules of ethanol (CJ 3CH2OH). 2.

Question

1.Calculate the mass in grams of 3.37x10^25 molecules of ethanol (CJ 3CH2OH).

2. Vitamin C has the formula C6H8O6. How many atoms of oxygen are in 2.96 g of vitamin C?

3. If Iron and sulfur react to form iron (III) Sulfide, calculate the mass in grams of sulfur needed to completely react with 2.7 moles of iron.

4. A chemist ran the reaction CH3CH2Cl+ KOH->^CH3CH2OH+KCl by heating 47.2 g of CH3Ch2Cl with excess KOH and obtained a ield of 28.7 of CH3CH2OH. What was his percent yield?

Please show work! I just want to get these concepts down before the exam and am having trouble understanding. Thanks!

Explanation / Answer

1.Calculate the mass in grams of 3.37x10^25 molecules of ethanol

First, change all to moles

1 mol = 6.022*10^23 molecules

(3.37*10^25) / (6.022*10^23) = 55.96 mol

MW of Methanol = 46.1 g/mol

mass = mol*MW = 55.96*46.1 = 2579.8 grams of Ethanol are present

2. Vitamin C has the formula C6H8O6. How many atoms of oxygen are in 2.96 g of vitamin C?

m = 2.96 of Vit C

MW of Vit C = 176.12 g/mol

mol Vit C = mass/MW = 2.96/176.12 = 0.01681 mol of Vit C

there are 6 mol of O per mol of Vit C so

0.01681 mol of Vit C will have 6*0.01681 mol of O = 0.10086 mol of O

1 mol = 6.022*10^23 atoms

0.01681 --> 0.01681*6.022*10^23 = 1.01*10^22 atoms of O

3. If Iron and sulfur react to form iron (III) Sulfide, calculate the mass in grams of sulfur needed to completely react with 2.7 moles of iron.

Fe + S --> FeS

2.7 mol of Fe will need 1:1 2.7 mol of S

MW of = 32.1 g/mol

therefore we need

mass = mol*MW = 2.7*32.1 = 86.67 g of S

4. A chemist ran the reaction CH3CH2Cl+ KOH->^CH3CH2OH+KCl by heating 47.2 g of CH3Ch2Cl with excess KOH and obtained a ield of 28.7 of CH3CH2OH. What was his percent yield?

m = 47.2 g of CH3CH2Cl

excess of KOH

m = 28.7 g of CH3CH2OH

% yield = real/theoretical

MW of CH3CH2Cl = 64.5 g/mol

mol of CH3CH2Cl = mass/MW = 47.2/64.5 = 0.73

ratio is 1:1 therefore expect 0.73 mol of CH3CH2OH

MW of CH3CH2OH = 46.1 g/mol

mass of Ethanol = 0.73*46.1 = 33.65 g

Compare

% yield = 28.7/ 33.65 * 100 = 85%

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.