How many grams of magnesium oxide are formed when 13.5 L of oxygen gas, measured at STP, completely reacts with magnesium metal according to the following reaction?
i got .496 g
but i know that like way off
To determine the number of grams of magnesium oxide formed when oxygen gas reacts with magnesium metal, we first need to write and balance the chemical equation for the reaction.
The balanced equation for the reaction between magnesium and oxygen is:
2 Mg + O2 -> 2 MgO
From the equation, we can see that 2 moles of magnesium react with 1 mole of oxygen gas to produce 2 moles of magnesium oxide.
Given that the volume of oxygen gas is 13.5 L at STP (Standard Temperature and Pressure), we can convert it to moles using the ideal gas law.
1 mole of any gas at STP occupies 22.4 L. Therefore:
13.5 L O2 * (1 mole O2/22.4 L O2) = 0.602 moles of O2
Now, using the stoichiometry from the balanced equation, we can calculate the number of moles of magnesium oxide formed.
Since the mole ratio between oxygen gas and magnesium oxide is 1:2, we have:
0.602 moles O2 * (2 moles MgO/1 mole O2) = 1.204 moles MgO
Finally, we can convert moles of magnesium oxide to grams using its molar mass.
The molar mass of magnesium oxide (MgO) is:
(1 mol Mg * molar mass Mg) + (1 mol O * molar mass O)
= (24.31 g/mol + 16.00 g/mol)
= 40.31 g/mol
Therefore:
1.204 moles MgO * (40.31 g MgO/1 mole MgO) = 48.51 g MgO
So, the correct answer is approximately 48.51 grams of magnesium oxide formed.
To answer this question, we need to use the stoichiometry of the given chemical reaction. The balanced equation for the reaction between magnesium metal (Mg) and oxygen gas (O2) is:
2 Mg + O2 -> 2 MgO
This equation tells us that 2 moles of magnesium react with 1 mole of oxygen gas to produce 2 moles of magnesium oxide.
Given that the volume of oxygen gas is 13.5 L at STP (Standard Temperature and Pressure), we need to convert this volume to moles. At STP, one mole of any ideal gas occupies 22.4 L.
So, using the ideal gas law, we can calculate the number of moles of oxygen gas:
moles of O2 = volume of O2 (L) / 22.4
moles of O2 = 13.5 L / 22.4 = 0.604 moles of O2
Now, using the stoichiometry of the reaction, we can calculate the number of moles of magnesium oxide formed. Since the ratio of moles of magnesium oxide produced to moles of oxygen gas consumed is 2:1, the moles of magnesium oxide formed will be half of the moles of oxygen gas.
moles of MgO = 0.604 moles of O2 / 2 = 0.302 moles of MgO
Finally, we can convert the moles of magnesium oxide to grams using the molar mass of magnesium oxide (MgO). The molar mass of magnesium is 24.3 g/mol, and that of oxygen is 16.0 g/mol.
molar mass of MgO = 24.3 g/mol (Mg) + 16.0 g/mol (O) = 40.3 g/mol
grams of MgO = moles of MgO × molar mass of MgO
grams of MgO = 0.302 moles of MgO × 40.3 g/mol = 12.17 g
Therefore, the correct answer is approximately 12.17 grams of magnesium oxide formed when 13.5 L of oxygen gas, measured at STP, completely reacts with magnesium metal.