Question

You mix 2.58 grams of NaHCO3 with 49.00 grams of 1.00M HCl. The two react to form aqueous NaCl, water, and carbon dioxide. The initial temperature was 21.4°C and temperature at the end of the reaction was 17.4°C. Assume that the density of HCl is 1.00g/mL.

a. Write the balanced reaction.
b. The final temperature of the resulting solution was lower than the initial temperature. How did the energy flow?
The energy was released by the solution and absorbed by the reaction.
The energy was absorbed by the solution and released by the reaction.

C. Use the equation q = nCΔT to calculate the heat (kJ) involved in the reaction.

D. Calculate the number of moles of sodium bicarbonate and hydrochloric acid used in the reaction.

Answer 1

This problem is providing the masses of both NaHCO3 and HCl and the initial and final temperature due to the reaction. It asks for the balanced chemical equation, the direction and quantity of heat and the consumed moles.

Calorimetry in chemical reactions.

In chemistry, calorimetry in chemical reactions basically allows us to quantify the chemical heat that is involved in a chemical change due to the rearrangement of bonds when the reactants get in contact.

In such a way, one can start this problem by writing and balancing the undergoing chemical equation between NaHCO3 and HCl, according to the law of conservation of mass:

$NaHCO_3 + HCl\rightarrow NaCl+H_2O+CO_2$

Which is already balanced as equal number of sodium, hydrogen, carbon and oxygen are present on both sides of the equation.

Next, since the final temperature of the resulting solution was lower than the initial temperature, one comes to the conclusion that the reaction is endothermic, as the energy was released by the solution and absorbed by the reaction.

Thereafter, given the general equation for the calculation of heat:

$Q=mC\Delta T$

One can calculate the heat involved in the reaction, assuming the specific heat of the solution equals waters:

$Q=-(2.58g+49.00g)*4.184\frac{J}{g\°C}(17.4-21.4)\°C \\\\Q=-863J*\frac{1kJ}{1000J}=0.863 kJ$

Which adopts the negative sign at the beginning, because of the flow of heat which make the heat of reaction the negative of that of the solution.

Finally, we use some stoichiometry to calculate the number of moles of sodium bicarbonate and hydrochloric acid used in the reaction:

$2.58gNaHCO_3 *\frac{1molNaHCO_3}{84.0gNaHCO_3}=0.0478molNaHCO_3\\ \\49.00gHCl*\frac{1molHCl}{36.46gHCl}=1.344molHCl$

Hence, since they react in a 1:1 mole ratio, we can see the excess HCl in the reaction, and thereby, the correct number of moles, reacting in the chemical reaction are just 0.0478 mol of the both of them.

Summary of answers:

a. $NaHCO_3 + HCl\rightarrow NaCl+H_2O+CO_2$

b. The reaction is endothermic, as the energy was released by the solution and absorbed by the reaction

c. 0.863 kJ

d. 0.0478 mol

Learn more about calorimetry: brainly.com/question/1407669