2Mg + O2 �¨ 2MgO
How many grams of MgO are produced when 79.5 grams of O2 react completely with Mg according to this equation?
To find the number of grams of MgO produced, we need to use stoichiometry, which relates the number of moles of reactants to the number of moles of products in a balanced chemical equation. Here's how we can solve this:
1. Begin by balancing the equation:
2Mg + O2 → 2MgO
2. Determine the molar masses of the elements involved:
Mg = 24.31 g/mol
O = 16.00 g/mol
3. Convert the given mass of O2 to moles:
Given mass = 79.5 g
Molar mass of O2 = 2 * 16.00 g/mol = 32.00 g/mol
Number of moles = Given mass / Molar mass = 79.5 g / 32.00 g/mol
4. Use the stoichiometric ratio from the balanced equation to find the moles of MgO produced:
According to the balanced equation, for every 1 mole of O2, 2 moles of MgO are produced.
Number of moles of MgO = (Number of moles of O2) * (2 moles of MgO / 1 mole of O2)
5. Calculate the mass of MgO produced:
Mass of MgO = Number of moles of MgO * Molar mass of MgO
Now let's calculate it step by step:
Number of moles of O2 = 79.5 g / 32.00 g/mol = 2.4844 mol
Number of moles of MgO = 2.4844 mol * (2 mol MgO / 1 mol O2) = 4.9688 mol
Mass of MgO = 4.9688 mol * (24.31 g/mol + 16.00 g/mol) = 204.6992 g
Therefore, when 79.5 grams of O2 react completely with Mg, 204.70 grams of MgO are produced.