Using the Stoichiometric Information Table below, if 9.00 g of \'D\' are obtaine

Question

Using the Stoichiometric Information Table below, if 9.00 g of 'D' are obtained, what is the percent yield? (Recall that "n" is an abbreviation for "mol". So "g/n" = "g/mol").

What is the concentration (weight % in water) of concentrated hydrobromic acid?

a 36%

b 100%

55%

48%

Given the following stoichiometric information for a generic reaction and that B is the limiting reagent, calculate the theoretical yield of F in moles.

a 36%

b 100%

c

55%

d

48%

Reaction A + 2B C + 2D Amounts 25.0g 15.0 MW (n) 111 gn 40 gn 74g/n 58.5g/m

Explanation / Answer

25 g of A + 15 g of B = 9 g of D

mol of A = mass/MW =25/111 = 0.2252

mol of B = mass/MW = 15/40 = 0.375

mol of D = mass/MW = 9/58.5 = 0.15384

note that ratio is 1:2 with respect to A and B so..

0.375 mol of B limits reaction, should produce --> 0.375 mol of D

then..

0.15384 mol of D required --> 1:2 --> 1/2*0.15384 = 0.07692 mol of A

0.15384 mol of D required --> 2:2 --> 0.15384 mol of B

%yield = D/theoretical * 100% = 0.15384/0.375 *100 = 41.024 % yield

What is the concentration (weight % in water) of concentrated hydrobromic acid?

The highest concentration possible for HBr ( undiluted) will be 48%, no more mass of HBr will "fit" per unit volume of liter

Given the following stoichiometric information for a generic reaction and that B is the limiting reagent, calculate the theoretical yield of F in moles.

if B limits, then

1 mol of B = 2 mol of F

0.6 mol of B --> 2*0.6 = 1.2 mol of F

the theoritcal yield will be 1.2 mol of F

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