In a reaction involving the iodination of acetone, following volumes were used t
ID: 484226 • Letter: I
Question
In a reaction involving the iodination of acetone, following volumes were used to make on mixture: 10 mL 4.0 M acetone + 10 mL 1.0 M HCI + 10 mL 0.0050 M I_2 + 20 mL H_2O How many moles of acetone were in the reaction mixture? Recall that, for a component A, no. moles A = M_A times V, where M_A is the molarity of A and V the volume in liters of the solution of A that was used. What was the molarity of acetone in the reaction mixture? The volume of the mixture was 50 mL, 0.050 L, and the number of moles of acetone was found in Part a. Again, M_a = no. moles A/V of soln. in liters How could you double the molarity of the acetone in the reaction mixture, keeping the total volume at 50 mL and keeping the same concentration of H^+ ion and I_2 as in the original mixture? Using the reaction mixture in Problem 1, a student found that it took 250 seconds for the color of the I_2 to disappear. What was the rate of the reaction? Given the rate from Part a, and the initial concentrations of acetone, H^+ ion, and I_2 in the reaction mixture, write Equation 3 as it would apply to the mixture. What are the unknowns that remain in the equation in Part b?Explanation / Answer
In a reaction involving the iodination of acetone, the following volumes were used to make up the reaction mixture:
10 ml 4.0 M acetone + 10 ml 1.0 M HCI + 10 ml 0.0050 M I2 + 20 ml H2O
A) How many moles of acetone were in the reaction mixture? Recall that, for a component A, no. moles A=Ma* V, where Ma is the molarity of A and V is the volume in liters of the solution of A that was used. ______________moles acetone
A)10 ml of 4 mole /1000ml soln
so you used
10 ml = .01 L
4.0M = 4 mol/L
10ml*(4mole/1000ml)=0.04 moles-------------answer
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B) What was the molarity of acetone in the reaction mixture? The volume of the mixture was 50 ml, 0.0050 L, and the number of moles of acetone was found in part a. again.
M= no. moles A/ V of soln. in _________________M acetone
=moles used/volume solution
= 0.04moles/(10ml + 10 ml + 10ml + 20 ml)*1000ml/liter
= 0.8M
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C) How could you double the molarity of the acetone in the reaction mixture, keeping the total volume at 50 ml and keeping the same concentration of H+ ion and I2 as in the original mixture?
) double the initial acetone volume to 20 ml and cut the water back to 10 ml
double the initial acetone to 20 ml ===> 0.08 moles acetone
then
molatity of final soln =
=0.08/(20+10+10+10ml)+1000ml/liter = 1.6 M
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