consider 2Mg+O2---> 2MgO
If 10.0g of o2 react, how much MgO will be formed?
mols O2 = grams/molar mass = ?
Convert mols O2 to mols MgO using the coefficients in the balanced equation.
Now convert mols MgO to grams; g - mols x molar mass = ?
To determine how much MgO will be formed when 10.0g of O2 reacts, we need to use stoichiometry and the molar ratios from the balanced chemical equation.
The balanced equation is:
2Mg + O2 → 2MgO
Step 1: Calculate the number of moles of O2.
To do this, we use the formula:
moles = mass / molar mass
The molar mass of O2 is 32.00 g/mol (16.00 g/mol for each O atom), so the number of moles of O2 is:
moles of O2 = 10.0 g / 32.00 g/mol = 0.3125 mol (rounded to 4 decimal places)
Step 2: Use the stoichiometric ratio to determine the number of moles of MgO produced.
From the balanced equation, we can see that the ratio of MgO to O2 is 2:1. This means that for every 2 moles of MgO, 1 mole of O2 is required.
Using this ratio, we can calculate the number of moles of MgO formed:
moles of MgO = (0.3125 mol O2) * (2 mol MgO / 1 mol O2)
moles of MgO = 0.625 mol (rounded to 3 decimal places)
Step 3: Calculate the mass of MgO formed.
To find the mass of MgO formed, we multiply the number of moles of MgO by its molar mass.
The molar mass of MgO is 40.31 g/mol (24.31 g/mol for Mg + 16.00 g/mol for O), so the mass of MgO formed is:
mass of MgO = 0.625 mol * 40.31 g/mol = 25.19 g (rounded to 2 decimal places)
Therefore, when 10.0g of O2 reacts, approximately 25.19g of MgO will be formed.