In general, if a reaction is spontaneous, the reactants possess more free energy than the products.
<u>TRUE </u>
FALSE
Cao + H2O ---->Ca(OH)2
Calculate the number of each reactant and the moles of the product
that is
moles = mass/molar mass
The moles of CaO= 56.08g/ 56.08g/mol(molar mass of Cao)= 1mole
the moles of water= 36.04 g/18 g/mol= 2.002moles
The moles of Ca (OH)2=74.10g/74.093g/mol= 1mole
The mass of differences of reactant and product can be therefore
explained as
1 mole of Cao reacted completely with 1 mole H2O to produce 1 mole of Ca(OH)2. The mass of water was in excess while that of CaO was limited
1. HCl (H:1, Cl 35.5) ---> 1+35.5 = 36.5
2. FeS (Fe:56, S:32) ---> 56+32 = 88
3. Cl2 (Cl:35.5) ---> 35.5 x 2 = 71
4. CaC03 (Ca:40, C: 12, O: 16) ---> 40 + 12 + 3(16) = 100
5. Fe0 (Fe: 56, O: 16) ---> 56+16=72
As long as the equation in question can be expressed as the sum of the three equations with known enthalpy change, its 
can be determined with the Hess's Law. The key is to find the appropriate coefficient for each of the given equations.
Let the three equations with given be denoted as (1), (2), (3), and the last equation (4). Let 
, 
, and 
be letters such that 
. This relationship shall hold for all chemicals involved.
There are three unknowns; it would thus take at least three equations to find their values. Species present on both sides of the equation would cancel out. Thus, let coefficients on the reactant side be positive and those on the product side be negative, such that duplicates would cancel out arithmetically. For instance, 
shall resemble the number of 
left on the product side when the second equation is directly added to the third. Similarly
Thus
and
Verify this conclusion against a fourth species involved- 
for instance. Nitrogen isn't present in the net equation. The sum of its coefficient shall, therefore, be zero.
Apply the Hess's Law based on the coefficients to find the enthalpy change of the last equation.
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