A student dissolves 12.1 g of potassium chloride (KCl) in 250. g of water in a well-insulated open cup. He then observes the temperature of the water fall from 21.0 °C to 17.1 °C over the course of 6.9 minutes. Use this data, and any information you need from the ALEKS Data resource, to answer the questions below about this reaction: KCl(s) → K (aq) + Cl (aq)

a. Is this reaction exothermic, endothermic, or neither?
b. Calculate the reaction enthalpy ΔH. rxn per mole of KCI.

a. Endothermic

b. 26.37kJ/mol

Explanation:

a. As we can see, the temperature of the water is decreasing when the reaction is occurring, that means the reaction is absorbing heat and is endothermic

b. To find the enthalpy we must find the change in heat when 12.1g of KCl are dissolved. Using the equation:

Q = -m*ΔT*C

Where Q is change in heat

m the mass of solution (250g + 12.1g = 262.1g)

ΔT is change in heat (17.1°C - 21.0°C = -3.9°C)

And C is specific heat of the solution (4.184J/g°C assuming is the same than the specific heat of water).

Replacing:

Q = -262.1g*-3.9°C*4.184J/g°C

Q = 4277J = 4.28kJ

As reaction enthalpy is the change in heat per mole of reaction, we must find the moles of 12.1g of KCl:

Moles KCl -Molar mass: 74.55g/mol-:

12.1g KCl * (1 mol / 74.55g) = 0.1623 moles KCl

The reaction enthalpy us:

4.28kJ / 0.1623mol = }

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