Answer:
[tex]X(16)=25.71grams[/tex]
Step-by-step explanation:
let [tex]X(t)[/tex] denote grams of [tex]C[/tex] formed in [tex]t[/tex] mins.
For [tex]X grams[/tex] of [tex]C[/tex] we have:
[tex]\frac{2}{3}Xg[/tex] of A and [tex]\frac{1}{3}Xg[/tex] of B
Amounts of A,B remaining at any given time is expressed as:
[tex]40-\frac{2}{3}Xg[/tex] of A and [tex]50-\frac{1}{3}Xg[/tex] of B
Rate at which C is formed satisfies:
[tex]\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}[/tex]
Apply the initial condition,[tex]X(0)=0[/tex] ,to the expression above
[tex]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}[/tex]
Now at [tex]X(8)=15[/tex]:
[tex]15=90-\frac{90}{90\times 8k+1} \ ->75=\frac{90}{720k+1}\\k=0.0002778[/tex]
Substitute in X(t) to get
[tex]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[/tex]
