75. zinc metal reacts with hydrochloric acid according to the following balanced equation: zn(s) + 2 hcl(aq)…

75. zinc metal reacts with hydrochloric acid according to the following balanced equation: zn(s) + 2 hcl(aq) → zncl₂(aq) + h₂(g) when 0.103 g of zn(s) is combined with enough hcl to make 50.0 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 δᵣh for this reaction as written. (use 1.0 gml⁻¹ for the density of the solution and 4.184 j g⁻¹°c⁻¹ as the specific heat capacity.)

75. zinc metal reacts with hydrochloric acid according to the following balanced equation: zn(s) + 2 hcl(aq) → zncl₂(aq) + h₂(g) when 0.103 g of zn(s) is combined with enough hcl to make 50.0 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 δᵣh for this reaction as written. (use 1.0 gml⁻¹ for the density of the solution and 4.184 j g⁻¹°c⁻¹ as the specific heat capacity.)

Answer

Explanation:

Step1: Calculate the mass of the solution

The volume of the solution $V = 50.0\ mL$, and the density of the solution $\rho=1.0\ g/mL$. Using the formula $m=\rho V$, we have $m = 1.0\ g/mL\times50.0\ mL=50.0\ g$.

Step2: Calculate the heat absorbed by the solution

The formula for heat $q = mc\Delta T$, where $m = 50.0\ g$, $c = 4.184\ J\ g^{-1}\ ^{\circ}C^{-1}$, and $\Delta T=T_{final}-T_{initial}=23.7^{\circ}C - 22.5^{\circ}C = 1.2^{\circ}C$. So $q=50.0\ g\times4.184\ J\ g^{-1}\ ^{\circ}C^{-1}\times1.2^{\circ}C = 251.04\ J$.

Step3: Calculate the number of moles of Zn

The molar - mass of $Zn$ is $M_{Zn}=65.38\ g/mol$. The mass of $Zn$ is $m_{Zn}=0.103\ g$. Using the formula $n=\frac{m}{M}$, we get $n_{Zn}=\frac{0.103\ g}{65.38\ g/mol}=0.001575\ mol$.

Step4: Calculate $\Delta_{r}H$

The heat released by the reaction is equal in magnitude but opposite in sign to the heat absorbed by the solution. For the reaction as written, the number of moles of $Zn$ in the balanced equation is 1 mole. We use the proportion $\frac{\Delta_{r}H}{1\ mol}=\frac{-q}{n_{Zn}}$. Substituting the values, $\Delta_{r}H=\frac{- 251.04\ J}{0.001575\ mol}\approx - 159390\ J/mol=-159.4\ kJ/mol$.

Answer:

$-159.4\ kJ/mol$