How many grams of Mg are needed to produce 224 g of MgO in the complete reaction of Mg with oxygen gas?

134.4 g of Mg

Explanation:

reaction:

2Mg + O2 ➡️ 2MgO

1) find the mol of MgO

mol = mass / molar mass

mass = 224 g

molar mass = 24+16 = 40

mol = 224 / 40

= 5.6 moles

2 mol = 5.6 moles

2) find the mass of Mg

mass = mol × molar mass

mol = 5.6

molar mass = 24

mass = 5.6 × 24

= 134.4 g

in the given question, the partial pressure is calculated by using the ideal gas equation, that is, pv = nrt

here, p is pressure, v is volume = 25 l, n is number of moles = 0.0104 moles, r is gas constant = -0.0821 l.atm / mol.k

t is temperature = 273 k

p = nrt / v

p = (0.0104 mol × 0.0821 l.atm / mol.k × 273 k) / 25 l

= 0.0093 atm