Given the mass (in grams) of a reactant A, first divide this by the molar mass of A to get the number of moles of A. Then balance the chemical reaction, and divide the number of moles of A by the coefficient of A in the balanced reaction. Next, find the coefficient of the product (B) in the reaction, and multiply this by the result from the previous operation. Finally, multiply by the molar mass of B to get the final answer in grams of the product B.
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Answer:
Any Alkali Earth Metal - Group 2A aka Group 16 or polyatomic cation with a 2+ charge
Answer:
25.97oC
Explanation:
Heat lost by aluminum = heat gained by water
M(Al) x C(Al) x [ Temp(Al) – Temp(Al+H2O) ] = M(H2O) x C(H2O) x [ Temp(Al+H2O) – Temp(H2O) ]
Where M(Al) = 23.5g, C(Al) = specific heat capacity of aluminum = 0.900J/goC, Temp(Al) = 65.9oC, Temp(Al+H2O)= temperature of water and aluminum at equilibrium = ?, M(H2O) = 55.0g, C(H2O)= specific heat capacity of liquid water = 4.186J/goC
Let Temp(Al+H2O) = X
23.5 x 0.900 x (65.9-X) = 55.0 x 4.186 x (X-22.3)
21.15(65.9-X) = 230.23(X-22.3)
1393.785 - 21.15X = 230.23X – 5134.129
230.23X + 21.15X = 1393.785 + 5134.129
251.38X = 6527.909
X = 6527.909/251.38
X = 25.97oC
So, the final temperature of the water and aluminum is = 25.97oC
