Start by taking a look at the balanced chemical equation
#color(red)(2)Al_((s)) + Fe_2O_(3(s)) -> 2Fe_((s)) + Al_2O_(3(s))#
The enthalpy change for this reaction is #DeltaH_"rxn" = "-852 kJ"#.
Notice that you have #color(red)(2)# moles of aluminium that react in order for the reaction to produce that much heat. In other words, if fewer moles of aluminium react, the reaction will produce less heat. Likewise, if more moles of aluminium react, the reaction will give off more heat.
To determine how many moles of aluminium react, use the element's molar mass
#30.5cancel("g") * "1 mole Al"/(26.98cancel("g")) = "1.1305 moles Al"#
This means that the reaction will give off
#1.1305cancel("moles Al") * (-"852 kJ")/(color(red)(2)cancel("moles Al")) = -"481.6 kJ"#
You know that all this energy went into heating #2.50* 10^(3)"g"# of water. The equation that links heat absorbed and change in temperature looks like this
#q = m * c * DeltaT#, where
#q# - the heat absorbed by the water;
#m# - the mass of the water;
#c# - the specific heat of water, equal to #4.18"J"/(g^@"C")#;
#DeltaT# - the change in temperature, defined as #T_"final" - T_"initial"#.
So, plug in your values into the equation and solve for #T_"final"#
#"481.6 kJ" = 2.50 * 10^(3)cancel("g") * 4.18"J"/(cancel("g") ^@cancel("C")) * (T_"final" - 20.2)^@cancel("C")#
#481.6 * cancel(10^3)cancel("J") = 2.50 * cancel(10^(3)) * 4.18cancel("J") * T_"final" - 211.09#
#T_"final" = (481.6 + 211.09)/(10.45) = 66.286""^@"C"#
Rounded to three sig figs, the answer will be
#T_"final" = color(green)(66.3""^@"C")#
