Hello. Could anyone help me out with a question I'm stuck on? Thanks in advance.(adsbygoogle = window.adsbygoogle || []).push({});

Calcium metal reacts with hydrochloric acid in a calorimeter and the enthalpy change is determined. Similarly, the enthalpy change for the reaction of calcium oxide with hydrochloric acid is determined. What is the enthalpy change for the following reaction:

[tex] Ca_{(s)}+ 1/2 O_{2} \longrightarrow CaO_{(s)} [/tex]

Evidence:

[tex] Ca_{(s)} + 2HCl_{(aq)} \longrightarrow CaCl_{2}_{(aq)} + H_{2}_{(g)} ,\triangle H_{r}=? [/tex]

[tex] CaO_{(s)} + 2HCl_{(aq)} \longrightarrow CaCl_{2}_{(aq)} + H_{2}O_{(l)} ,\triangle H_{r}=? [/tex]

Temperature Change Due to Reaction With 100 mL of 1.0 mol/L HCl

Reactant: Ca(s)

Mass(g) : 0.52

Initial Temp: 21.3 °C

Final Temp: 34.5 °C

Reactant: CaO(s)

Mass (g): 1.47

Initial Temp: 21.1 °C

Final Temp: 28.0 °C

So far I have:

for Ca:

[tex] nH= mc\triangle t [/tex]

[tex] H= \frac{mc\triangle t}{n} [/tex]

H= {(0.100 L)*(1.0 mol/1 L)*(0.03646 kg/1 mol)*(4.19 kJ/kg°C)*(34.5°C-21.3°C)} / { (0.52g)*(1 mol/40.08 g) } = - 15.54 kJ/mol

for CaO:

H= {(0.100 L)*(1.0 mol/1 L)*(0.03646 kg/1 mol)*(4.19 kJ/kg°C)*(28.0°C-21.1°C)} / { (1.47g)*(1 mol/56.08 g) } = -4.02 kJ/mol

Not too sure where to go from here but the answers are suppose to be -0.43 MJ,+ 0.11 MJ and -0.60 MJ,also I'm not too sure if that's the order the answers appear either.

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# Homework Help: Enthalpy Change

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