Question

"Magnesium metal reacts quantitatively with oxygen to give magnesium oxide. If 5.00 g of Mg and 5.00 g of O2 are allowed to react, what mass of MgO is formed? What amount (in grams) of the excess reactant is left over at the end?"

A) 10.00 g of MgO, 0.00 g of excess Mg
B) 10.00 g of MgO, 0.50 g of excess O2
C) 7.33 g of MgO, 0.67 g of excess O2
D) 7.33 g of MgO, 0.67 g of excess Mg

Answer 1

In this reaction, magnesium metal (Mg) reacts with oxygen (O2) to form magnesium oxide (MgO). To find the mass of MgO formed and the amount of excess reactant remaining, we compare the moles of reactants and use the mole ratio in the balanced equation.

Step-by-step explanation:

In this reaction, magnesium metal (Mg) reacts with oxygen (O2) to form magnesium oxide (MgO). To determine the mass of MgO formed, we need to find the limiting reactant and the excess reactant.

First, we calculate the moles of Mg and O2 using their molar masses (24.31 g/mol for Mg and 32.00 g/mol for O2). Then, we compare the mole ratio of Mg to O2 in the balanced equation (2:1) to determine which reactant is limiting.

Since the reaction uses a 1:1 mole ratio of Mg to MgO, we can use the moles of Mg to determine the moles of MgO formed. Finally, we multiply the moles of MgO by its molar mass (40.31 g/mol) to find the mass of MgO formed.

To calculate the mass of the excess reactant remaining, we subtract the moles of O2 consumed from the initial moles of O2 given and multiply by its molar mass (32.00 g/mol).