$2018/19$ Question Number $(2)$ Workthrough

a$)$

b$)$ Using $C_3{H}_8+S{O}_{2}3C{O}_2+4H_{2}O($Assume $100$ mol of flue gas$)20kg?$

Thus, $\%$ excess of $O_2$ air is: $\frac{\frac{300}{103}-\frac{2270}{1133}}{\frac{2270}{1133}}100\%$~~$45\mathrm{.}37\%($too big!$)$ Alternatively: $\frac{\frac{25}{11}-\frac{2270}{1133}}{\frac{220}{1133}}100\%$~~$\frac{\text{A}\text{c}\text{t}\text{u}\text{a}\text{l}\text{~}\text{~}28\mathrm{.}1\%}{13\mathrm{.}44\%(too\text{s}\text{m}\text{a}\text{l}\text{l}\text{!})}$