Can someone help me with QUESTION 3C...the last question ANSWERS TO THE QUESTION

Question

Can someone help me with QUESTION 3C...the last question

ANSWERS TO THE QUESTIONS ABOVE

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) ??????????

Assume the reaction reached equilibrium (c) Calculate Keq 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. ANSWERS TO THE QUESTIONS ABOVE 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) ??????????

Explanation / Answer

Keq=[AB2]/[A][B]^2

molarity=no of moles/total volme,L(total volume is40+50ml=90ml=0.09L)

no of moles of A left unreacted=1.17*10^-4/0.09L=0.0013M

no of moles of B left unreacted=3.2*10^-4/0.09L=0.0035M

no of moles of AB2 formed=0.000395/0.09L=0.00438M

Keq=0.00438M/(0.0013Mx0.0035M^2)=275039.2

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