For the reaction, calculate how many grams of the product form when 3.2 g of Mg completely reacts.
Assume that there is more than enough of the other reactant.
2Mg(s)+O2(g)→2MgO(s)
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To find the grams of the product formed when 3.2 g of Mg completely reacts, we need to use stoichiometry and the balanced equation.
Step 1: Write down the balanced equation:
2Mg(s) + O2(g) -> 2MgO(s)
This equation tells us that 2 moles of Mg react with 1 mole of O2 to produce 2 moles of MgO.
Step 2: Calculate the molar mass of Mg:
The molar mass of Mg is approximately 24.31 g/mol.
Step 3: Convert grams of Mg to moles:
To do this, divide the given mass of Mg by its molar mass:
Number of moles = mass / molar mass
Number of moles of Mg = 3.2 g / 24.31 g/mol ≈ 0.1319 mol
Step 4: Use stoichiometry to find the moles of MgO produced:
From the balanced equation, we can see that 2 moles of Mg produce 2 moles of MgO. Therefore, the moles of MgO produced will be the same as the moles of Mg, since both reactants have a stoichiometric coefficient of 2.
Number of moles of MgO = 0.1319 mol
Step 5: Convert moles of MgO to grams:
To do this, multiply the number of moles by the molar mass of MgO:
Mass of MgO = moles × molar mass
Mass of MgO = 0.1319 mol × (24.31 g/mol + 16.00 g/mol) ≈ 5.07 g
Therefore, when 3.2 g of Mg completely reacts, approximately 5.07 grams of MgO will be formed.