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 40 grams of A and 50 grams of B, and for each gram of B, 2 grams of A is used. It is observed that 20 grams of C is formed in 10 minutes. How much (in grams) is formed in 20 minutes? (Round your answer to one decimal place.)

In each 3 grams of C, there are 2 grams of A and 1 gram of B. So, for some amount C, the amount remaining of A is 40 -(2C/3), and the amount remaining of B is (50 -C/3). Since the reaction rate is proportional to the product of these amounts, we have ...

C' = k(40 -2C/3)(50 -C/3) = (2k/9)(60 -C)(150 -C) . . . for some constant k

This is separable differential equation with a solution of the form ...

ln((150 -C)/(60 -C)) = at + b

We know that C(0) = 0, so b=ln(150/60) = ln(2.5). And we know that C(10) = 20, so ln(130/40) = 10a +ln(2.5) ⇒ a = ln(1.3)/10

Then our equation for C is ...

ln((150 -C)/(60 -C)) = t·ln(1.3)/10 +ln(2.5)

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For t=20, this is ...

ln((150 -C)/(60 -C)) = 2ln(1.3) +ln(2.5) = ln(2.5·1.3²) = ln(4.225)

Taking antilogs, we have ...

(150 -C)/(60 -C) = 4.225

1 +90/(60 -C) = 4.225

C = 60 -90/3.225 ≈ 32.093 . . . . . grams of product in 20 minutes

In 20 minutes, about 32.1 grams of C are formed.