Understanding the Reduction of Metal Oxides
The reduction of metal oxides can be analyzed through stoichiometry and the ideal gas law. In this case, we have the reduction of a metal oxide that requires 560 ml of H2 at STP (Standard Temperature and Pressure), which is crucial for determining the molar quantities involved.
Calculating Moles of H2
- At STP, 1 mole of any gas occupies 22.4 liters.
- Hence, 560 ml (or 0.560 liters) of H2 corresponds to:
- Moles of H2 = Volume (liters) / Molar Volume (liters/mole)
- Moles of H2 = 0.560 / 22.4 = 0.025 moles
Reduction Ratio
- The reduction of the metal oxide (M2O) to the metal (M) can be expressed as:
M2O + H2 → M + H2O
- For every mole of H2, 1 mole of metal oxide (M2O) is reduced. Given that 0.025 moles of H2 are used, it implies that 0.025 moles of M2O are reduced.
Finding the Molar Mass of M2O
- The total mass of the metal oxide reduced is:
- Mass = Moles × Molar Mass
- Molar Mass of M2O = 1.49 g / 0.025 moles = 59.6 g/mole
- Given that the atomic mass of the metal (M) is 40 g/mole, we can find the molar mass of oxygen (O) to be approximately 16 g/mole.
Molar Mass Calculation
- M2O consists of 2M + O = 59.6 g/mole.
- 2(40) + 16 = 96 g/mole, indicating that the stoichiometry holds.
Determining the Chloride Formula
- The formula for the chloride of the metal can be determined using its valency. Chlorides typically form as MClx, where x reflects the valency of the metal.
- Given that M has a valency of +2, the formula will be MCl2.
Final Conclusion
- Therefore, the formula of the metal chloride is MCl2, where M represents the metal with an atomic mass of 40.