Please work problems step by step, so I can understand. Question 1 CH 3 COOH + H

Question

Please work problems step by step, so I can understand.

Question 1

CH3COOH + HOCH2CH3? CH3COOCH2CH3 + H2O

Based on the equation above, what mass of acetic acid (CH3COOH) is needed to prepare 252 g of ethyl acetate (CH3COOCH2CH3) if the percent yield of the reaction is 85%?

(Assume ethanol is present in excess, ethyl acetate = 88.676 g/mol, and acetic acid = 60.0536 g/mol.)

Question 1 options:

296 g

300 g

252 g

201 g

Question 2

If the actual yield of Mg3N2 is 47.45 g and the theoretical is 51.09 g. What is the percent yield?

Question 3

Calculate the mass percent of carbon in one mole of the following compound, if the molecular mass of the compound is 74.12 g.

C4H9OH

Question 4

Question 4 options:Calculate the mass percent of P in the following if the molecular mass of the compound is 263.9 g. (Report you answer using the proper significant digits.)

Al = 26.9815 g/mol

P = 30.9738 g/mol

O = 15.9994 g/mol

296 g

300 g

252 g

201 g

Explanation / Answer

Ans 1 : 201 g

Percent yield = (actual yield / theoretical yield) x 100

85 = (252 / m) x 100

m = 296.47 g

Number of moles of ethyl acetate = 296.47 / 88.676 = 3.34 mol

So number of mol of acetic acid required will also be 3.34 mol

Mass of acid required = 3.34 x 60.0536

= 201 g

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