This is a standard calorimetry problem.
The basic principle is that all the heats involved must add up to zero.
#"Heat of reaction" + "Heat to warm water" = 0#
#q_1 +q_2 = 0#
#nΔH_"rxn" + mcΔT# = 0
#n = 1.86 color(red)(cancel(color(black)("g MgO"))) × (1"mol MgO")/(40.30 color(red)(cancel(color(black)("g MgO")))) = "0.046 15 mol MgO"#
#m = 100.0 color(red)(cancel(color(black)("mL"))) × "1.000 g"/(1color(red)(cancel(color(black)("mL")))) = "100.0 g"#
#ΔT = "(35.7 – 21.3) K" = "14.4 K"#
∴ #"0.046 15 mol" ×ΔH_"rxn" + 100.0 color(red)(cancel(color(black)("g"))) × "4.18 J·"color(red)(cancel(color(black)("K"^"-1""g"^"-1"))) × 14.4 color(red)(cancel(color(black)("K"))) = 0#
#"0.046 15 mol" × ΔH_"rxn" + "6019 J" = 0#
#ΔH_"rxn" = -"6019 J"/"0.046 15 mol" = "-130 423 J/mol" = "-130.4 kJ/mol"#
Here's a video on how to use calorimetry to calculate enthalpies of reaction.
