(i) The rate of formation of X is proportional to the product of the masses of Y and Z, which are \(10-x\) and \(20-x\) respectively. Therefore, \(\frac{dx}{dt} = k(10-x)(20-x)\). Given \(\frac{dx}{dt} = 2\) when t = 0 and x = 0, we have \(2 = k(10)(20)\), so \(k = 0.01\). Thus, \(\frac{dx}{dt} = 0.01(10-x)(20-x)\).

(ii) Separate variables: \(\frac{1}{(10-x)(20-x)} dx = 0.01 dt\).

Use partial fractions: \(\frac{1}{(10-x)(20-x)} = \frac{A}{10-x} + \frac{B}{20-x}\).

Solving gives \(A = \frac{1}{10}\) and \(B = -\frac{1}{10}\).

Integrate: \(-\frac{1}{10} \ln(10-x) + \frac{1}{10} \ln(20-x) = 0.01t + C\).

At t = 0, x = 0: \(C = \frac{1}{10} \ln 2\).

Rearrange: \(-\ln(10-x) + \ln(20-x) = 0.1t + \ln 2\).

\(\ln\left(\frac{20-x}{10-x}\right) = 0.1t + \ln 2\).

\(\frac{20-x}{10-x} = 2\exp(0.1t)\).

\(20-x = 2(10-x)\exp(0.1t)\).

\(x = \frac{20(\exp(0.1t)-1)}{2\exp(0.1t)-1}\).

(iii) As t becomes large, \(\exp(0.1t)\) becomes very large, so \(x\) approaches 10.