What is the enthalpy change when 5.2 mol of lead(II) oxide reacts with excess sulfur dioxide
2PbO(s) + 2SO2(g) ===> 2PbS(s) + 3O2 +827.4 kJ
5.2molPbO x +827.4kJ/2molPbO = 4302 kJ
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2PbO(s) + 2SO2(g) ===> 2PbS(s) + 3O2 +827.4 kJ
5.2molPbO x +827.4kJ/2molPbO = 4302 kJ