Question

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 15 grams of C is formed in 8 minutes. How much (in grams) is formed in 16 minutes? (Round your answer to one decimal place.)

Answer 1

Answer:

$X(16)=25.71grams$

Step-by-step explanation:

let $X(t)$ denote grams of $C$ formed in  $t$ mins.

For $X grams$ of $C$ we have:

$\frac{2}{3}Xg$ of A and $\frac{1}{3}Xg$ of B

Amounts of A,B remaining at any given time is expressed as:

$40-\frac{2}{3}Xg$ of A and  $50-\frac{1}{3}Xg$  of B

Rate at which C is formed satisfies:

$\frac{dX}{dt} \infty(40-\frac{2}{3}X)(50-\frac{1}{3}X)->\frac{dX}{dt}=k(90-X)\\\therefore \frac{dX}{(90-X)^2}=kdt->\int{\frac{dX}{(90-X)^2}} \, =\int {k} \, dt \\\therefore \frac{1}{90-X}=kt+c->90-X=\frac{1}{kt+c}\\\\X(t)=90-\frac{1}{kt+c}$

Apply the initial condition,$X(0)=0$ ,to the expression above

$0=90-\frac{1}{c} \ \ ->c=\frac{1}{90}\\\therefore\\X(t)=90-\frac{1}{kt+\frac{1}{90}} \ \ ->X(t)=90-\frac{90}{90kt+c}$

Now at $X(8)=15$:

$15=90-\frac{90}{90\times 8k+1} \ ->75=\frac{90}{720k+1}\\k=0.0002778$

Substitute  in X(t) to get

$X(t)=90-\frac{90}{0.0002778t\times 90+1}\\X(t)=90-\frac{90}{0.25t+1}\\But \ t=16\\\therefore X(t)=90-\frac{90}{0.025\times16+1}\\X(t)=25.71$