What is the mass of bromine in 11.2g of MgBr2?
1 answer:
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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