Answer
$m_{sulfur}=0.014\frac{kg}{h}$
$m_{H_{2}SO_{3}}=0.0359\frac{kg}{h}$
Work Step by Step
First we calculate the rate of fuel entering the engine:
$m_{fuel}=\frac{m_{air}}{ratio_{air-fuel}}=\frac{336\frac{kg}{h}}{18\frac{kg_{air}}{kg_{fuel}}}=18.67\frac{kg}{h}$
And the mass of sulfur in this fuel will be:
$m_{sulfur}=18.67\frac{kg}{h}*750ppm=0.014\frac{kg}{h}$
Knowing this molar masses:
H: $2*1\frac{kg}{kmol}=2\frac{kg}{kmol}$
S: $1*32\frac{kg}{kmol}=32\frac{kg}{kmol}$
O: $3*16\frac{kg}{kmol}=48\frac{kg}{kmol}$
Then the molar mass of the sulfurous acid is:
$M_{H_{2}SO_{3}}=2\frac{kg}{kmol}+32\frac{kg}{kmol}+48\frac{kg}{kmol}=82\frac{kg}{kmol}$
And the flow rate of sulfurous acid added to the environment for each kmol is:
$m_{H_{2}SO_{3}}=\frac{M_{H_{2}SO_{3}}}{M_{sulfur}}m_{sulfur}=\frac{82\frac{kg}{kmol}}{32\frac{kg}{kmol}}*0.014\frac{kg}{h}=0.0359\frac{kg}{h}$
