Westonci.ca is your go-to source for answers, with a community ready to provide accurate and timely information. Join our Q&A platform to get precise answers from experts in diverse fields and enhance your understanding. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform.

Zinc metal reacts with HCl according to the balanced equation:Zn + 2HCl à ZnCl2 + H2When 0.103 g of Zn is combined with enough HCl to make 50 mL of solution in a coffee-cup calorimeter, all of the zinc reacts, raising the temperature of the solution from 22.5 C to 23.7 C. Find the heat of the reaction. (ans: -160 kJ/mol)

Sagot :

Answer:

-163kJ/mol

Explanations:

Given the reaction between the zinc metal reacts with HCl according to the balanced equation:

[tex]Zn+2\text{HCl}\rightarrow\text{ZnCl}_2+H_2[/tex]

The required heat of reaction will be calculated using the formula below:

[tex]\triangle H_{\text{rxn}}=-\frac{Q}{n_{Zn}}[/tex]

where:

• Q is the ,heat produced

,

• nZn is the, number of moles that reacted

The formula for calculating the quantity of heat produced is expressed as:

[tex]Q=mc\triangle\theta[/tex]

where:

• m is the ,mass, of the ,Zinc metal

,

• c is the ,specific heat capacity, of zinc

,

• △θ is th,e change in temperature

Get the mass of zinc

[tex]\begin{gathered} \text{mass}=\text{ }density\times\text{volume} \\ m=\rho\times v \end{gathered}[/tex]

The quantity of heat becomes:

[tex]Q=\rho\cdot v\cdot C\triangle\theta[/tex]

Substitute the given parameters to have:

[tex]\begin{gathered} Q=\frac{1.02\cancel{g}}{\cancel{mL}}\times50\cancel{mL}\times\frac{4.18J}{\cancel{g^oC^{}}}\times(23.7-22.5)\cancel{^oC} \\ Q=1.02\times50\times4.18\times1.2 \\ Q=255.8Joules \end{gathered}[/tex]

Next is to get the number of moles of Zinc that reacted (nZn)

[tex]\begin{gathered} nZ_n=\frac{Mass}{\text{Molar mass}} \\ nZ_n=\frac{0.103g}{65.4g\text{/mol}} \\ nZ_n=0.00157mole \end{gathered}[/tex]

Get the required heat of reaction of Zinc:

[tex]\begin{gathered} \triangle H_{\text{rxn}}=-\frac{Q}{nZ_n} \\ \triangle H_{\text{rxn}}=-\frac{255.8Joules}{0.00157moles} \\ \triangle H_{\text{rxn}}=-162,929.93J\text{/mol} \\ \triangle H_{\text{rxn}}\approx-163kJ\text{/mol} \end{gathered}[/tex]

Hence the heat of the reaction of Zinc is approximately -163kJ/mol