Ca(OH)2(aq) + 2HCl(aq)------> CaCl2(aq) + 2H2O(l) ΔH-?
CaO(s) + 2HCl(aq)-----> CaCl2(aq) + H2O(l), Δ<span>H = -186 kJ
</span>
CaO(s) + H2O(l) -----> Ca(OH)2(s), Δ<span>H = -65.1 kJ
</span>
1) Ca(OH)2 should be reactant, so
CaO(s) + H2O(l) -----> Ca(OH)2(s)
we are going to take as
Ca(OH)2(s)---->CaO(s) + H2O(l), and ΔH = 65.1 kJ
2) Add 2 following equations
Ca(OH)2(s)---->CaO(s) + H2O(l), and ΔH = 65.1 kJ
<span><u>CaO(s) + 2HCl(aq)-----> CaCl2(aq) + H2O(l), and ΔH = -186 kJ</u>
</span>Ca(OH)2(s)+CaO(s) + 2HCl(aq)--->CaO(s) + H2O(l)+CaCl2(aq) + H2O(l)
Ca(OH)2(s)+ 2HCl(aq)---> H2O(l)+CaCl2(aq) + H2O(l)
By addig these 2 equation, we got the equation that we are needed,
so to find enthalpy of the reaction, we need to add enthalpies of reactions we added.
ΔH=65.1 - 186 ≈ -121 kJ
The empirical formula : MnO₂.
<h3>Further explanation</h3>
Given
632mg of manganese(Mn) = 0.632 g
368mg of oxygen(O) = 0.368 g
M Mn = 55
M O = 16
Required
The empirical formula
Solution
You didn't include the pictures, but the steps for finding the empirical formula are generally the same
- Find mol(mass : atomic mass)
Mn : 0.632 : 55 = 0.0115
O : 0.368 : 16 =0.023
- Divide by the smallest mol(Mn=0.0115)
Mn : O =
The empirical formula : MnO₂
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