Mg reaction with O₂ gas will produce MgO so the equation will be
2Mg+O₂⇒2MgO. (You have to find the equation in order two figure out the number of moles of O₂ that will react with 1 mole of MgO).
The first step is to find the number of moles of Mg in 4.03g of Mg. You can do this by dividing 4.03g Mg by its molar mass (which is 24.3g/mol) to get 0.1658mol Mg. Then you have to find the number of moles of O₂ that will react with 0.1658mol Mg. To do this you need to use the fact that 1mol O₂ will react with 2mol Mg (this reatio is from the chemical equation) so you have to multiply 0.1658mol Mg by (1mol O₂)/(2mol Mg) to get 0.0829mol O₂. From here you would usually use PV=nRT and solve for V However, the question tells us that we are at STP, that means you can use the fact that 22.4L of gas is 1 mol of gas at STP. Using that information we can find the volume of O₂ gas by mulitlying 0.0829mol O₂ by 22.4L/mol to get 1.857L which equals 1857mL.
therefore, 1857mL of O₂ gas will react with 4.03g of Mg.
I hope this helps. Let me know in the comments if anything is unclear.
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
<span>Kind of substance besides water:
The best example of hydrogen bonding excluding water is DNA. The two strands of polymers are connected by hydrogen bonds between the nucleotide bases</span>.
Answer: Option B. 76.83L
Explanation:
1 mole of a gas occupy 22.4L at stp. This implies that 1mole of Radon also occupy 22.4L at stp.
If 1 mole of Radon = 22.4L
Therefore, 3.43 moles of Radon = 3.43 x 22.4 = 76.83L
Venus so that the sun rises in the west
