在某一特定的化学反应中，每 2 克 $U$ 与 1 克 $V$ 结合并转化为 3 克 $W$。设 $u$、$v$ 和 $w$ 分别表示在时间 $t$（以分钟为单位）时存在的 $U$、$V$ 和 $W$ 的质量（以克为单位）。根据质量作用定律，$w$ 对 $t$ 的变化率与 $u$ 和 $v$ 的乘积成正比。初始时，$u=40$，$v=50$ 且 $w=0$。

1. Express the remaining masses of $U$ and $V$ in terms of $w$.

Since 2 grams of $U$ and 1 gram of $V$ produce 3 grams of $W$, the masses of $U$ and $V$ consumed to produce $w$ grams of $W$ are $\frac{2}{3}w$ and $\frac{1}{3}w$ respectively.
Thus, the remaining masses at time $t$ are:
$u = 40 - \frac{2}{3}w$
$v = 50 - \frac{1}{3}w$

2. Use the law of mass action to write the differential equation.

The rate of change of $w$ is proportional to $uv$. Let $k_1$ be the constant of proportionality:
$\frac{dw}{dt} = k_1 u v = k_1 \left(40 - \frac{2}{3}w\right)\left(50 - \frac{1}{3}w\right)$

3. Factorise the expression into the required form and show that $k > 0$.

$\frac{dw}{dt} = k_1 \left[ -\frac{2}{3}(w - 60) \right] \left[ -\frac{1}{3}(w - 150) \right]$
$\frac{dw}{dt} = \frac{2}{9}k_1 (w - 60)(w - 150)$
Let $k = \frac{2}{9}k_1$, giving $\frac{dw}{dt} = k(w - 60)(w - 150)$.
Since $W$ is being formed initially, the rate of change is positive ($\frac{dw}{dt} > 0$) when $w=0$.
$k(0 - 60)(0 - 150) > 0 \implies 9000k > 0 \implies k > 0$
Hence shown.

1. Separate variables and apply partial fractions to prepare for integration.

$\frac{dw}{dt} = k(w-60)(w-150)$
$\int \frac{1}{(w-60)(w-150)} \,dw = \int k \,dt$
Using partial fractions:
$\frac{1}{(w-60)(w-150)} = \frac{A}{w-60} + \frac{B}{w-150}$
$1 = A(w-150) + B(w-60)$
When $w = 60$, $-90A = 1 \implies A = -\frac{1}{90}$
When $w = 150$, $90B = 1 \implies B = \frac{1}{90}$
$\frac{1}{90} \int \left( \frac{1}{w-150} - \frac{1}{w-60} \right) \,dw = \int k \,dt$

2. Integrate both sides to obtain the general solution.

$\frac{1}{90} (\ln|w-150| - \ln|w-60|) = kt + C$
$\frac{1}{90} \ln\left|\frac{w-150}{w-60}\right| = kt + C$
Since $w$ starts at $0$ and must be less than $60$ (the limiting mass), $\frac{w-150}{w-60} > 0$. We can rewrite it without the modulus sign as:
$\frac{1}{90} \ln\left(\frac{150-w}{60-w}\right) = kt + C$

3. Use the initial condition to find the constant of integration $C$.

Given that initially $w = 0$ when $t = 0$:
$\frac{1}{90} \ln\left(\frac{150-0}{60-0}\right) = k(0) + C$
$C = \frac{1}{90} \ln\left(\frac{150}{60}\right) = \frac{1}{90} \ln(2.5)$
Substituting $C$ back into the equation:
$\frac{1}{90} \ln\left(\frac{150-w}{60-w}\right) = kt + \frac{1}{90} \ln(2.5)$
$\ln\left(\frac{150-w}{60-w}\right) = 90kt + \ln(2.5)$

4. Apply the condition at $t = 5$ to evaluate the parameter $k$.

When $t = 5$, $w = 10$:
$\ln\left(\frac{150-10}{60-10}\right) = 90k(5) + \ln(2.5)$
$\ln\left(\frac{140}{50}\right) = 450k + \ln(2.5)$
$450k = \ln(2.8) - \ln(2.5) = \ln\left(\frac{2.8}{2.5}\right) = \ln(1.12)$

5. Substitute $t = 20$ into the specific equation to solve for $w$.

When $t = 20$:
$\ln\left(\frac{150-w}{60-w}\right) = 90k(20) + \ln(2.5)$
$\ln\left(\frac{150-w}{60-w}\right) = 4(450k) + \ln(2.5)$
$\ln\left(\frac{150-w}{60-w}\right) = 4\ln(1.12) + \ln(2.5)$
$\ln\left(\frac{150-w}{60-w}\right) = \ln(1.12^4 \times 2.5)$
$\frac{150-w}{60-w} = 1.12^4 \times 2.5 \approx 3.9338$
$150 - w = 3.9338(60 - w)$
$150 - w = 236.0279 - 3.9338w$
$2.9338w = 86.0279$
$w \approx 29.323$
$w = 29.32 \text{ (to 2 d.p.)}$

1. Identify the behaviour of the solution's exponential term as $t \to \infty$.

From the integration in part (ii), the relationship between $w$ and $t$ can be expressed as:
$\frac{150-w}{60-w} = 2.5 e^{90kt}$
Since $k$ is a positive constant, as $t \to \infty$, the exponent $90kt \to \infty$, and therefore $e^{90kt} \to \infty$.
(Alternatively, if $w$ was made the subject: $w = \frac{150 - 150e^{-90kt}}{2.5 - e^{-90kt}}$, as $t \to \infty$, $e^{-90kt} \to 0$)

2. Deduce the limiting value of $w$.

For $\frac{150-w}{60-w}$ to approach $\infty$, the denominator must approach $0$:
$60 - w \to 0$
$w \to 60$
(From the alternative expression: $w \to \frac{150 - 0}{2.5 - 0} = 60$)

Thus, as $t$ becomes very large, the mass $w$ approaches $60$ grams.