How many grams of MgO are produced when 40.0grams of 02 react with Mg?
2mg+O2--->2mgO
Could you please set the problem up?
I think the answer is 101, but I do not know how to get that answer. Thank you
To calculate the mass of MgO produced when 40.0 grams of O2 reacts with Mg, we need to use the stoichiometry of the balanced chemical equation provided.
The balanced equation is: 2Mg + O2 -> 2MgO
The molar mass of O2 is 32.00 g/mol and the molar mass of MgO is 40.31 g/mol.
To set up the problem, we follow these steps:
Step 1: Convert the given mass of O2 to moles.
Given Mass of O2 = 40.0 g
Molar mass of O2 = 32.00 g/mol
Moles of O2 = Given Mass / Molar mass = 40.0 g / 32.00 g/mol = 1.25 mol
Step 2: Use stoichiometry to find the moles of MgO produced.
From the balanced equation, we see that 1 mol of O2 reacts with 2 mol of MgO.
So, Moles of MgO = Moles of O2 x (2 mol MgO / 1 mol O2) = 1.25 mol x 2 = 2.50 mol
Step 3: Convert moles of MgO to grams.
Moles of MgO = 2.50 mol
Molar mass of MgO = 40.31 g/mol
Mass of MgO = Moles x Molar mass = 2.50 mol x 40.31 g/mol = 100.77 g
Therefore, when 40.0 grams of O2 react with Mg, 100.77 grams of MgO are produced. The answer is approximately 100.77 grams, not 101 grams as you mentioned earlier.
To find the number of grams of MgO produced when 40.0 grams of O2 react with Mg, we need to use the given balanced chemical equation and the concept of stoichiometry.
The balanced chemical equation is: 2Mg + O2 -> 2MgO
First, let's calculate the number of moles of O2 using the given mass. We can do this using the molar mass of O2, which is 32.00 g/mol:
moles of O2 = mass of O2 / molar mass of O2
moles of O2 = 40.0 g / 32.00 g/mol
moles of O2 = 1.25 mol
According to the balanced equation, the reaction requires 1 mole of O2 to produce 2 moles of MgO. Therefore, the number of moles of MgO produced will be twice the moles of O2:
moles of MgO = 2 * moles of O2
moles of MgO = 2 * 1.25 mol
moles of MgO = 2.50 mol
Finally, to find the mass of MgO produced, we can multiply the number of moles of MgO by its molar mass, which is 40.30 g/mol:
mass of MgO = moles of MgO * molar mass of MgO
mass of MgO = 2.50 mol * 40.30 g/mol
mass of MgO = 101.25 g
Therefore, 40.0 grams of O2 would produce 101.25 grams of MgO. So your answer of 101 grams is close, but the actual value is slightly higher due to more accurate calculations.
Here is a worked example that will work all of your stoichiometry problems.
http://www.chemteam.info/SigFigs/SigFigs.html