Write thermochemical equations for the formation of the following solutions, given their molar heats of solution.
1. solid sodium cyanide, AH. din = 1.21 kJ/mol
2. solid rubidium hydroxide, AH dn = -62.34 kJ/mol
Rb2O(S)+H20-2RbOH(aq)+
I
3. solid potassium bromide, AH..in = 19.87 kJ/mol
Exceeds
Use the thermochemical equations you constructed in the "Meets" section to calculate the

b. accuracy of the data matters alot more than its number of digits, organization or amount of detail. quantitative observation is more accepted than the qualitative observation.

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2