Someone please help me it’s an assessment on Stochiometry

Chemistry

1 answer:

HACTEHA [7]4 years ago

6 0

Answer:

2) 0.3787moles

3) 198g

4) 7.658g

5) 12.26g

Explanation:

2C6H6 + 9O2 -> 12CO2 + 6H20

Weight of a CO2 Mol 12+ 16x2 = 44g

In 100g number of moles = 100/44= 2.2727 Moles

Number of C6H6 2moles burned when creating 12 moles of CO2

Therefore 2.2727/12 x 2 = 0.3787

3H2SO4 + 2Al(OH)3 -> Al2(SO4)3 + 6H2O

H2O weight of a 1mol is = 2+16 = 18g

2moles of Al(OH)3 cerates 6 moles of water.

3moles of sulfuric creates 6 moles of water.

5.5moles Al(OH)3 reacts with 5.5 moles of sulfuric

Therefore number of water moles = 5.5/3 x 6 = 11moles

Weight water = 11 x 18 = 198g

Cu + 2AgNO3 -> 2Ag + Cu(NO3)2

1mol of Cu produces 2moles of silver

Ag = 107.86 g/Mol

Cu = 63.54 g/Mol

In 26g Ag number of moles = (26/107.86)/2 x 63.54 = 7.658g

3K2CrO4 + 2AlPO4 -> 2K3PO4 + Al2(CrO4)3

K2CrO4 = 39x2 + 52 + 16x4 = 194g/Mol

Al2(CrO4)3 = 54 + 354 = 408g/Mol

3moles of K2CrO4 creates 1mol of Al2(CrO4)3

Therefore weight of the Al2(CrO4)3 produced by 17.5g potassium chromate

(17.5/194)/3 moles of Al2(CrO4)3 produced = 0.03moles

Al2(CrO4)3 produced = 0.03 x 408 = 12.26g

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) Predict the product for the following reaction sequence. A) 2,2-dimethylpropanoic acid B) 3,3-dimethyl-2-butanone C) 2,2-dimet

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Answer:

Option C. 2,2-dimethyl-1-propanol

Explanation:

The first attached picture, is the initial reactant.

Now, what happens in each step?, at first we have an alkyl bromide reacting with Mg/ether. This converts the reactant in a grignard reagent, which is often used in several ketones/ladehydes reactions, and even some acid and esthers.

Now, in the second step, we see this grignard reagent, reacting with CO₂ and H₃O⁺. At first, the grignard reagent will attach to the oxygen of the CO₂ by the MgBr, and the carbonated chain will attach to the carbonile. Then in acid medium, the OMGBr leaves the molecule, and form a carboxylic acid.

Finally, in the last step with LiAlH₄ and water, this compound just reduces the carboxylic acid to a primary alcohol, therefore, the only choice available is option c, 2,2-dimethyl-1-propanol.

See picture 2 below for the drawing.

Hope this helps