If 22.4 dm3 of steam react with MgO, calculate the mass of the product that will form.

To calculate the mass of the product that will form when 22.4 dm3 of steam reacts with MgO, we need to follow a few steps:

1. Find the balanced chemical equation for the reaction between steam (H2O) and MgO. The balanced equation is:
MgO + H2O -> Mg(OH)2

2. Determine the molar ratio between MgO and the product Mg(OH)2. From the balanced equation, we can see that the stoichiometric ratio is 1:1. This means that for every 1 mole of MgO, we will obtain 1 mole of Mg(OH)2.

3. Convert the given volume of steam (22.4 dm3 or liters) to moles. To do this, we need to 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 ideal gas constant, and T is the temperature.

Assuming the reaction takes place at standard temperature and pressure (STP) conditions (0 degrees Celsius and 1 atm), we can use the molar volume of a gas at these conditions, which is 22.4 L/mol.

Therefore, 22.4 dm3 of water vapor is equivalent to 22.4 L or 1 mole of H2O.

4. Since the stoichiometric ratio between MgO and Mg(OH)2 is 1:1, the number of moles of Mg(OH)2 produced will also be 1.

5. Finally, use the molar mass of Mg(OH)2 to calculate the mass of the product. The molar mass of Mg(OH)2 is the sum of the atomic masses of Mg, O, and 2x H.

The atomic masses are as follows:
Mg: 24.31 g/mol
O: 16 g/mol
H: 1.008 g/mol

Therefore, the molar mass of Mg(OH)2 is:
(24.31 + 16 + 2x1.008) g/mol

Multiply the molar mass by the number of moles obtained in step 4 to calculate the mass of the product formed.