24.3 g of magnesium yield 40.3 g MgO upon combustion with oxygen. How much magnesium needs to be burned to yield 12.1 g MgO?
Please help! I can't figure this out. Little guidance here!
To solve this problem, we need to use the concept of stoichiometry, which is the relationship between the amounts of reactants and products in a chemical reaction.
First, we need to determine the molar masses of magnesium (Mg) and magnesium oxide (MgO). The molar mass of Mg is 24.31 g/mol, and the molar mass of MgO is 40.31 g/mol.
Next, we need to calculate the number of moles of MgO produced when 24.3 g of Mg reacts. We can use the equation:
Moles of substance = Mass of substance / Molar mass of substance
Moles of MgO = 40.3 g / 40.31 g/mol = 1 mol MgO
Now, we need to use the balanced chemical equation to determine the stoichiometric ratio between Mg and MgO. The balanced equation for the combustion of magnesium is:
2 Mg + O2 → 2 MgO
From the equation, we can see that 2 moles of Mg react to produce 2 moles of MgO. Therefore, the stoichiometric ratio between Mg and MgO is 2:2, or 1:1.
Since we know that 1 mole of Mg reacts to produce 1 mole of MgO, we can conclude that 24.3 g of Mg reacts to produce 40.3 g of MgO. Therefore, the molar ratio is:
24.3 g Mg / 40.3 g MgO = 1 mol Mg / 1 mol MgO
Now, we can use this molar ratio to calculate how much magnesium needs to be burned to yield 12.1 g of MgO.
1 mol Mg × (12.1 g MgO / 40.31 g MgO) = 0.3 mol Mg
Finally, we need to convert the moles of Mg to grams:
0.3 mol Mg × 24.31 g/mol = 7.29 g Mg
Therefore, to yield 12.1 g of MgO, you will need to burn 7.29 g of magnesium.
24.3/40.3 = x/12.1
trick question, no need to calculate mols or anything, just proportion.