Question

i just need help on this first question, the others parts are just added information if its needed

1. First, calculate the amount of heat energy, Ical, produced in each reaction using Eq. 1 from the Introduction. Convert Vwater to Mwater using a density value of 1.00 g/mL, and s = 4.18 J/(g•°C). Next, calculate the number of moles of NaNO3 (and NH4Cl) present in each reaction using Eq. 2. Finally, calculate the enthalpy change, AH, for each reaction in terms of kJ/mol of each reactant. And check your signs, because 9system = - Ycal! SHOW ALL CALCULATIONS. 1 qsystem = - 9cal = - S.Msolution AT,

Answer 1

ANSWER:

Data:

• Volume of water in each trial = 100.0 mL
• Density of water = 1.00 g / mL
• Specific heat of solution = 4.18 J / g.ºC

	Trial 1 (NaNO3)	Trial 2 (NH4Cl)
Mass of ionic solid	8.5 g	5.4 g
Temperature change (ΔT)	- 4.2 ºC	- 3.2 ºC

First we calculate the mass of water:

$mass\, H_{2}O=100.0\, mL\times \frac{1.00\, g}{1\, mL}=100.0\, g$

Now, we calculate the mass of solution for both solutions

• NaNO3 solution

• NH4Cl solution

mass NH4Cl(aq) = mass H2O + mass NH4CIS)

Then we calculate the energy produced in each reaction (qsystem)

$q_{system}=-q_{cal}=-s\times mass_{(solution)}\times \Delta T$

• NaNO3 solution

$q_{NaNO_{3}(aq)}=-q_{cal}=-4.18 \frac{J}{g. ^{\circ}C} \times 108.5\, g \times (-4.2\, ^{\circ}C)$

• NH4Cl solution

$q_{NH_{4}Cl(aq)}=-q_{cal}=-4.18 \frac{J}{g. ^{\circ}C} \times 105.4\, g \times (-3.2\, ^{\circ}C)$

We need to calculate the moles of NaNO3 and NH4Cl

• NaNO3 (MW = 84.99 g/mol)

$moles\, NaNO_{3}=8.5\, g \, NaNO_{3} \times \frac{1\, mol\, NaNO_{3}}{84.99\, g\, NaNO_{3}} =0.100\, mol$

• NH4Cl solution

$moles\, NH_{4}Cl=5.4\, g \, NH_{4}Cl \times \frac{1\, mol\, NH_{4}Cl}{53.49\, g\, NH_{4}Cl} =0.101\, mol$

Finally, the ΔHRxn for each reaction is

• NaNO3 solution

$\Delta H_{Rxn}=\frac{q_{NaNO_{3}(aq)}}{moles\, NaNO_{3}}=\frac{1904.83\, J}{0.100\, mol}$

$\Delta H_{Rxn}=19048\, \frac{J}{mol}=19.048\, \frac{kJ}{mol}$

• NH4Cl solution

$\Delta H_{Rxn}=\frac{q_{NH_{4}Cl(aq)}}{moles\, NH_{4}Cl}=\frac{1409.83\, J}{0.101\, mol}$

$\Delta H_{Rxn}=13959\, \frac{J}{mol}=13.959\, \frac{kJ}{mol}$