Can someoen help me with this chemistry question. ANSWERS TO THE QUESTIONS: 1a)

Question

Can someoen help me with this chemistry question.

ANSWERS TO THE QUESTIONS:

1a) 0.000512

1b) 0.00111

2a) A

2b) 0.000512

2c) 77.1%

3a) 1.17*10^-4

3b) 3.2*10^-4

3c) ??????????

***PROFESSOR'S HINT*** :

The moles of AB2 formed (which is
needed to get its molarity) is simply the
number given in the opening paragraph
(do NOT use the answer to 2(b)). Try to
avoid rounding error by not truncating
your molarities when you calculate them.

Two chemicals A & B react according to the following equation: A(aq) +2B(aq) right arrow AB2(aq) Suppose that 40.0 mL of 0.0128 M A(aq) is mixed with 50.0 mL of 0.0222 M B(aq) and, after the reaction, 0.000395 moles of AB2(aq) were obtained. 1. (a) Calculate the moles of A initially present before the reaction. (b) Calculate the moles of B initially present before the reaction. 2. ASSUMING THE REACTION WENT TO COMPLETION: (a) Given the amounts of reactants that were mixed, determine which should be the limiting reactant A or B? (b) How many moles of 2 would theoretically have been produced? (c) What is the actual percent yield? 3. ASSUMING THE REACTION REACHED EQUILIBRIUM Using the moles of reactant initially present, and the moles of product that actually formed: (a) How many moles of A are left unreacted? (b) How many moles of B are left unreacted? (c) Calculate Keg for the reaction. To do this, three molarities will need to be calculated. To get these molarities, divide each number of moles (of A left, B left, and AB2 formed) by the total volume of 0.0900 L. Can someoen help me with this chemistry question. ANSWERS TO THE QUESTIONS: 1a) 0.000512 1b) 0.00111 2a) A 2b) 0.000512 2c) 77.1% 3a) 1.17*10^-4 3b) 3.2*10^-4 3c) ?????????? ***PROFESSOR'S HINT*** : The moles of AB2 formed (which is needed to get its molarity) is simply the number given in the opening paragraph (do NOT use the answer to 2(b)). Try to avoid rounding error by not truncating your molarities when you calculate them.

Explanation / Answer

1 (a) No of moles Of A = 0.04 L x 0.0128 = 0.000512 moles

(b) No of moles Of B= 0.05 L x 0.0222 = 0.00111 moles

2(a) As per the equation

1 mole of A requires moles of B = 2

0.000512 mole of A requires B = 0.001024 mole

2 moles of B requires moles of A = 1 mole

0.00111 mole of B requires A = 0.00111/2 =0.000555 moles

But we have only 0.000512 mole of A

Therefore A is limiting reactant

(b) As A is limiting reactant

Therefore

1mole of A gives mole of AB2 = 1 mole

0.000512 mole of A gives mole of AB2 = 0.000512 mole of AB2

(c) Percentage yield = actual yield x 100/theoretical yield = 0.000395 x100/0.000512 = 77.15 %

3(a) Moles of A left unreacted = 0.000512 - 0.000395 = 0.000117 mole of A

(b) Moles of B left unreacted = 0.00111-(0.000395x2)= 0.00032 mole of B

(c) Keq =[ AB2]/[A][B] = (0.000395/0.09)/( 0.00111/0.09)(0.000512/0.09) = 0.004/0.012 x 0.006 = 555.55

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