when heat gained = heat lost
when AL is lost heat and water gain heat
∴ (M*C*ΔT)AL = (M*C*ΔT) water
when M(Al) is the mass of Al= 225g
C(Al) is the specific heat of Al = 0.9
ΔT(Al) = (125.5 - Tf)
and Mw is mass of water = 500g
Cw is the specific heat of water = 4.81
ΔT = (Tf - 22.5)
so by substitution:
∴225* 0.9 * ( 125.5 - Tf) = 500 * 4.81 * (Tf-22.5)
∴Tf = 30.5 °C
Divide that my the molar mass which is 23 so 1.4087 g
Info: Al(oh)3 might be an improperly capitalized: Al(OH)3
Error: Some elements or groups in the reagents are not present in the products: O
Error: equation Al4C3+H2O=Al(oh)3+CH4 is an impossible reaction
Please correct your reaction or click on one of the suggestions below:
Al4C3 + H2O = Al(OH)3 + CH4
Heating up, mostly. Solid can be heated to liquid, then to gas, then eventually to plasma.
