Question

If 1.32 1.19 g MgO is combined with 100.0 mL of 1.00 M HCl in a coffee cup calorimeter, the temperature of the HCl solution increases from 24.2 C to 34.4 36.8C. Assume that the specific heat of the HCl solution is 4.18 J/g·°C and its density is 1.00 g/mL.

Answer 1

The reaction could be written as:

$MgO + 2HCl \rightarrow MgCl_{2}+H_{2}O;$

The amount in moles of MgO:

$n(MgO)=\frac{1.32g}{40g/mol}=0.033mol;$

The amount in moles of HCl:

$n(HCl)=0.100L*1.00mol/L=0.100mol;$

The mass of the reaction mixture:

$M(reaction mixture)=m(MgO)+V_{HCl}*d(HCl)=1.32g+100mL*1.00g/mL=101.32g;$

Thus, the amount of the heat released:

$Q=mC\delta{t}=101.32g*4.18\frac{J}{g*C}*(36.8C-24.2C)=5336J;$

The molar heat is:

$Q_{mol}=\frac{5336J}{0.033mol}=162\frac{kJ}{mol};$

In the case of 1.19 g og MgO the molar heat would be 144 kJ/mol.