CH4(g)+2O2 arrow CO2(g)+2H2O(1) delta H=-890.3kJ. Calculate delta Hwhen 5 grams of methane react with an excess of oxgen?o i just divide Delta H by 5?

To calculate the change in enthalpy (ΔH) for a specific amount of substance involved in a chemical reaction, you need to use stoichiometry and the given molar ratio.

Here's how you can calculate the ΔH when 5 grams of methane (CH4) react with an excess of oxygen (O2):

1. Determine the molar mass of CH4. The molar mass of carbon (C) is 12.01 g/mol, and the molar mass of hydrogen (H) is 1.01 g/mol. So, the molar mass of CH4 is (12.01 g/mol) + (4 × 1.01 g/mol) = 16.05 g/mol.

2. Convert the mass of CH4 to moles. Since the molar mass of CH4 is 16.05 g/mol, divide the given mass (5 g) by the molar mass to get moles: 5 g / 16.05 g/mol = 0.31 mol CH4.

3. Use the balanced chemical equation to determine the stoichiometry. According to the equation CH4(g) + 2O2(g) → CO2(g) + 2H2O(l), the molar ratio between CH4 and CO2 is 1:1. That means that for every 1 mole of CH4 reacted, 1 mole of CO2 is produced.

4. Calculate the ΔH. Since the ΔH given is for the reaction of 1 mole of CH4, we need to multiply it by the number of moles of CH4 involved in the reaction. In this case, ΔH = -890.3 kJ/mol × 0.31 mol CH4 = -275.7 kJ.

Therefore, the change in enthalpy (ΔH) when 5 grams of methane react with an excess of oxygen is -275.7 kJ. Note that the negative sign indicates that the reaction is exothermic (releasing heat).