There are two heat transfers to consider:
#"heat released by reaction + heat absorbed by water = 0"#
#color(white)(mmmmmm)q_1 color(white)(mmmmml)+color(white)(mmmmmm) q_2 color(white)(mmmml)= 0#
#color(white)(mmmmm)nΔ _text(r)H color(white)(mmmml)+ color(white)(mmmm)mC_text(s)ΔT color(white)(mmml)= 0#
Let's calculate each of these heats separately.
Step 1. Calculate #q_1#
#q_1 = nΔ_text(r)H = 0.110 color(red)(cancel(color(black)("mol"))) × ("-131 kJ"·color(red)(cancel(color(black)("mol"^"-1")))) = "-14.41 kJ" = "-14 410 J"#
Step 2. Calculate #q_2#
#m color(white)(l)= "75 g"#
#C_text(s) color(white)(l)= "4.184 J·°C"^"-1""g"^"-1"#
#ΔT = ?#
∴ #q_2 = 75 color(red)(cancel(color(black)("g"))) × "4.184 J·°C"^"-1"color(red)(cancel(color(black)("g"^"-1"))) × ΔT = 314 color(white)(l)ΔTcolor(white)(l) "J·°C"^"-1"#
Step 3. Calculate #T_text(f)#
#ΔT = T_text(f) - T_text(i)#
#q_1 + q_2 = 0#
#"-14 410" color(red)(cancel(color(black)("J"))) + 314 color(white)(l)ΔTcolor(red)(cancel(color(black)("J")))·"°C"^"-1" = 0#
#ΔT = "14 410"/("314 °C"^"-1") = "45.9 °C"#
#T_text(f) = T_text(i) + ΔT = "(25 + 45.9) °C = 71 °C"#
