I am having trouble deciding what formula to use for the following problem. Any help would be appreciated.
Magnesium in flashbulbs burns according to the equation
2 Mg (s) + O2 (g) --> 2MgO (s)
What mass of Mg combines with 5.80 L of O2 measured at STP?
Convert 5.80 L O2 to moles.
(1 mole of O2 at STP = 22.4 L)
Since 2 moles Mg combine with 1 mole O2,
moles of Mg = (2)(moles O2)
Grams Mg = (moles Mg)(24.3 g Mg/mol Mg)
Thank you!
what is the correct answer for this?
To solve this problem, you need to use the concept of stoichiometry, which involves calculating the amount of reactants and products in a chemical equation.
First, let's convert the volume of O2 from liters to moles. At Standard Temperature and Pressure (STP), the volume of 1 mole of any ideal gas is 22.4 liters.
Given:
Volume of O2 = 5.80 L
Step 1: Convert the volume of O2 to moles.
moles of O2 = volume of O2 / 22.4 L/mol
moles of O2 = 5.80 L / 22.4 L/mol
moles of O2 = 0.259 Moles
Now, according to the balanced equation, the stoichiometric ratio between Mg and O2 is 2:1. This means that for every 2 moles of Mg, we need 1 mole of O2. Since we have the number of moles of O2, we can calculate the number of moles of Mg that will react.
Step 2: Calculate the moles of Mg.
moles of Mg = (moles of O2) / 1
moles of Mg = 0.259 Moles / 1
moles of Mg = 0.259 Moles
Finally, since we know the molar mass of Mg, we can convert the moles of Mg to grams.
Step 3: Calculate the mass of Mg.
molar mass of Mg = 24.31 g/mol
mass of Mg = (moles of Mg) × (molar mass of Mg)
mass of Mg = 0.259 Moles × 24.31 g/mol
mass of Mg = 6.29 grams
Therefore, approximately 6.29 grams of Mg will combine with 5.80 liters of O2 at STP.