Final answer:
To find the number of moles of MgO formed from a given volume of oxygen gas, we use the stoichiometry of the reaction. 55 L of O2 is equal to 2.46 moles of O2, which would result in 4.92 moles of MgO being formed.
Step-by-step explanation:
To find the number of moles of MgO formed, we need to determine the stoichiometry of the reaction between oxygen gas (O2) and magnesium (Mg). The balanced chemical equation for this reaction is:
2Mg + O2 -> 2MgO
From the equation, we can see that 2 moles of Mg react with 1 mole of O2 to form 2 moles of MgO. Therefore, if 1 mole of O2 reacts, we would expect 2 moles of MgO to be formed.
Since 55 L of O2 is given, we need to convert the volume to moles using the ideal gas law. Assuming standard conditions (STP), 1 mole of any ideal gas occupies 22.4 L. Therefore, 55 L of O2 would be:
55 L * (1 mole / 22.4 L) = 2.46 moles of O2
Since 1 mole of O2 reacts to form 2 moles of MgO, we can determine the number of moles of MgO formed:
2.46 moles O2 * (2 moles MgO / 1 mole O2) = 4.92 moles of MgO
Therefore, the correct answer is B. 4.9