Answer:
Part A
The volume of the gaseous product is
Part B
The volume of the the engine’s gaseous exhaust is
Explanation:
Part A
From the question we are told that
The temperature is 
The pressure is 
The of 
The chemical equation for this combustion is

The number of moles of
that reacted is mathematically represented as

The molar mass of
is constant value which is
So 

The gaseous product in the reaction is
and water vapour
Now from the reaction
2 moles of
will react with 25 moles of
to give (16 + 18) moles of
and 
So
1 mole of
will react with 12.5 moles of
to give 17 moles of
and 
This implies that
0.8754 moles of
will react with (12.5 * 0.8754 ) moles of
to give (17 * 0.8754) of
and 
So the no of moles of gaseous product is


From the ideal gas law

making V the subject

Where R is the gas constant with a value 
Substituting values
Part B
From the reaction the number of moles of oxygen that reacted is


The volume is


No this volume is the 21% oxygen that reacted the 79% of air that did not react are the engine gaseous exhaust and this can be mathematically evaluated as

Substituting values