To determine the oxidation numbers of Al4C3, we first need to know the oxidation number of aluminum (Al) and carbon (C) in their respective compounds. Aluminum typically has an oxidation number of +3, while carbon has an oxidation number of -4 in carbides. Since there are 4 aluminum atoms in Al4C3, the total positive charge from aluminum is +12. To balance this, the total negative charge from carbon must be -12, which means each carbon atom has an oxidation number of -4. **

To determine the oxidation numbers of Al4C3, we first need to understand that the oxidation number of aluminum (Al) is typically +3 and the oxidation number of carbon (C) is typically -4 in carbides. Since there are four aluminum atoms in Al4C3, the total positive charge from aluminum is +12. The total negative charge from carbon is -12. To balance the charges, we can assign an oxidation number of +3 to each aluminum atom and -4 to each carbon atom in Al4C3. **

To determine the volume of CH4 obtained from 250 mg of Al4C3 at standard conditions, we first need to calculate the moles of Al4C3. The molar mass of Al4C3 is 144.0 g/mol. Therefore, 250 mg of Al4C3 is equal to 0.25 g, which is 0.00174 moles. From the balanced chemical equation for the reaction of Al4C3 with water, we know that 1 mole of Al4C3 produces 4 moles of CH4. Therefore, 0.00174 moles of Al4C3 will produce 0.00696 moles of CH4. Finally, using the ideal gas law at standard conditions (1 atm and 273 K), we can calculate the volume of CH4 obtained. **

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