Answer:
We can use the information given in the problem to calculate the RAM (relative atomic mass) of X and the minimum volume of hydrogen used in the reaction.
First, we need to calculate the mass of oxygen in the oxide XO:
mass of oxygen = mass of oxide - mass of metal
mass of oxygen = 4.5 g - 3.6 g
mass of oxygen = 0.9 g
Next, we can use the mass of oxygen to calculate the number of moles of oxygen:
moles of oxygen = mass of oxygen / molar mass of oxygen
moles of oxygen = 0.9 g / 16.00 g/mol
moles of oxygen = 0.05625 mol
Since the oxide XO is formed by the combination of X and oxygen, we can assume that the mass of X in the oxide is equal to the mass of the oxide minus the mass of oxygen:
mass of X = mass of oxide - mass of oxygen
mass of X = 4.5 g - 0.9 g
mass of X = 3.6 g
We can use the mass of X and the number of moles of oxygen to calculate the number of moles of X:
moles of X = mass of X / atomic mass of X
moles of X = 3.6 g / atomic mass of X
Combining this with the stoichiometry of the reaction, which tells us that 1 mole of XO reacts with 1 mole of H2 to produce 1 mole of X and 1 mole of H2O, we can write:
moles of H2 = moles of X / 1 = (3.6 g / atomic mass of X) / 1
To determine the minimum volume of hydrogen at STP (Standard Temperature and Pressure), we can use the ideal gas law:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles, R is the universal gas constant, and T is the temperature. At STP, the pressure and temperature are 1 atm and 273 K, respectively, and the universal gas constant is 0.08206 L atm/(mol K).
Substituting the given values and solving for V, we get:
V = nRT/P = [(3.6 g / atomic mass of X) / 1] * (0.08206 L atm/(mol K)) * 273 K / 1 atm
Simplifying and solving for the atomic mass of X, we get:
atomic mass of X = (3.6 g / V) * (1 mol/2 mol of H2) * (1 mol of XO/1 mol of X) * (16.00 g/mol of oxygen + atomic mass of X)
Substituting the given values, we get:
atomic mass of X = (3.6 g / V) * (0.5) * (1 / 1) * (16.00 g/mol + atomic mass of X)
Multiplying out, we get:
atomic mass of X = (1.8 g / V) * (16.00 g/mol + atomic mass of X)
Solving for atomic mass of X, we get:
atomic mass of X = (1.8 g / V) * 16.00 g/mol / (1 - 1.8 g / V)
Substituting V = 22.4 L (the volume of 1 mole of gas at STP), we get:
atomic mass of X = (1.8 g / 22.4 L) * 16.00 g/mol / (1 - 1.8 g / 22.4 L)
atomic mass of X ≈ 56.1 g/mol
Therefore, the RAM of X is approximately 56.1 g/mol, and the minimum volume of hydrogen used in the reaction is approximately 22.4 L at STP.