magnesium in flashbulbs burn according to the equation

2Mg + O2 ---> 2MgO

what mass of Mg combines with 5.80 L of O2 measured at STP?

How many liters pf O2 will combine with 3.5g at STP

To determine the mass of Mg that combines with 5.80 L of O2 measured at STP (Standard Temperature and Pressure), you need to follow these steps:

Step 1: Determine the number of moles of O2:
To calculate the number of moles of O2, you can use the ideal gas law equation:

PV = nRT

Where:
P = pressure in atmospheres (atm)
V = volume in liters (L)
n = number of moles
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature in Kelvin (K)

At STP (standard temperature and pressure), the values are:
P = 1 atm
T = 273.15 K

Substituting the known values:
V = 5.80 L
P = 1 atm
T = 273.15 K

Rearranging the ideal gas law equation to solve for n:
n = PV / RT

n = (1 atm * 5.80 L) / (0.0821 L·atm/(mol·K) * 273.15 K)

n ≈ 0.25 moles of O2.

Step 2: Use the balanced equation to determine the moles of Mg:
From the balanced equation:
2Mg + O2 ---> 2MgO

The stoichiometry is 2 moles of Mg reacts with 1 mole of O2. Therefore, if we have 0.25 moles of O2, we would need half that amount of Mg.

0.25 moles of O2 * (2 moles of Mg / 1 mole of O2) = 0.50 moles of Mg.

Step 3: Convert moles to grams:
To convert moles of Mg to grams, we use the molar mass of Mg, which is approximately 24.31 g/mol.

Mass of Mg = moles of Mg * molar mass of Mg
Mass of Mg = 0.50 moles * 24.31 g/mol
Mass of Mg ≈ 12.16 grams.

Therefore, the mass of Mg that combines with 5.80 L of O2 measured at STP is approximately 12.16 grams.