Two chemicals a and b are combined to form a chemical

c. the rate, or velocity, of the reaction is proportional to the product of the instantaneous amounts of a and b not converted to chemical

c. initially, there are 100 grams of a and 50 grams of b, and for each gram of b, 2 grams of a is used. it is observed that 15 grams of c is formed in 10 minutes. how much is formed in 40 minutes? (round your answer to one decimal place.) grams

-dB/dt = kAB = k (2B) (B) = 2kB^2

-dB/B^2 = 2kdt

Integrating: 1/B - 1 / (B_0) = 2kt

At t = 10, if 15 g of C have formed, this must have consumed 10 g A and 5 g B. The remaining mass of B is 45 g.

1/45 - 1/50 = 2 (k) (10)

k = 1.11 x 10^-4

Then substituting this value of k with t = 40:

1/B - 1/50 = 2 (1.11 x 10^-4) (40)

1/B - 1/50 = 0.008889

1/B = 0.028889

B = 34.62 g remaining

Therefore, 50 - 34.62 = 15.38 g of B have been consumed.

Doubling, 30.76 g of A have been consumed.

This means that 15.38 + 30.76 = 46.15 g of C have been formed.