For the reaction, calculate how many grams of the product form when 3.0 g of Mg completely reacts.

Assume that there is more than enough of the other reactant.
2Mg(s)+O2(g)→2MgO(s)

Isn't this very close to this problem I showed you earlier?

http://www.jiskha.com/display.cgi?id=1478745034

To calculate the amount of product formed in a chemical reaction, you need to use the stoichiometry of the balanced equation. In this case, the balanced equation is:

2Mg(s) + O2(g) → 2MgO(s)

The stoichiometry of the balanced equation tells us that 2 moles of Mg react with 1 mole of O2 to produce 2 moles of MgO.

To find the number of moles of Mg, we divide the given mass of Mg (3.0 g) by its molar mass. The molar mass of Mg is 24.31 g/mol.

Number of moles of Mg = Mass of Mg / Molar mass of Mg
Number of moles of Mg = 3.0 g / 24.31 g/mol

Now, we can use the stoichiometry from the balanced equation to calculate the number of moles of MgO produced.

According to the stoichiometry, 2 moles of Mg react to produce 2 moles of MgO.

Number of moles of MgO = Number of moles of Mg x (2 moles of MgO / 2 moles of Mg)
Number of moles of MgO = (3.0 g / 24.31 g/mol) x (2 mol/2 mol)

Finally, we can calculate the mass of MgO produced by multiplying the number of moles of MgO by its molar mass. The molar mass of MgO is 40.31 g/mol.

Mass of MgO = Number of moles of MgO x Molar mass of MgO
Mass of MgO = [(3.0 g / 24.31 g/mol) x (2 mol/2 mol)] x 40.31 g/mol

Calculate the expression to find the mass of the product.