In an experiment to determine the enthalpy change for this reaction, you combine 0.158 g of Mg metal with enough HCl to make 100.0 mL of solution in a coffee-cup calorimeter. The HCl is sufficiently concentrated so that the Mg completely reacts. The temperature of the solution rises from 25.6 °C to 32.8 °C as a result of the reaction. Find ΔHrxn for the reaction as written. Use 1.00 g/mL as the dens

The value of Δ for the reaction =  463

Explanation:

Δ = - ------ (1)

We know that

(1)(100) × 4.18 × (32.8 - 25.6)

 3010 J = 3.01 Kilo Joule

From equation (1)

Δ  = 3.01 Kilo Joule

No. of moles

m = 0.158 gm & M = 24.31  gm Mg

No. of moles    

N = 0.0065

Therefore

Δ  =

Δ  =  463

This is the value of Δ for the reaction.

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explanation:

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