A student dissolves 14.9 g of ammonium chloride (NH4Cl in 250. g of water in a well-insulated open cup. She then observes the temperature of the water fall from 20.0 °C to 16.0 °C over the course of 5.2 minutes. Use this data, and any information you need from the ALEKS Data resource, to answer the questions below about this reaction: NH, Cl(s) – NH (aq) + Cl (aq) You can make any reasonable assumptions about the physical properties of the solution. Be sure answers you calculate using measured data are rounded to 2 significant digits. Note for advanced students: it's possible the student did not do the experiment carefully, and the values you calculate may not be the same as the known and published values for this reaction. O exothermic Is this reaction exothermic, endothermic, or neither? endothermic x I ? neither If you said the reaction was exothermic or endothermic, calculate the amount of heat that was released or absorbed by the reaction in this case. kJ

Calculate the reaction enthalpy AH xn per mole of NHACI.

Endothermic

4.433kJ

15.91 kJ / mol

Explanation:

In water, ammonium chloride dissolves thus:

NH₄Cl(s) → NH₄⁺(aq) + Cl⁻(aq)

As the student observed the temperature decreased, the reaction is endothermic because absorbs heat from surroundigns when occurs.

It is possible to find the absorbed heat using coffee cup calorimeter equation:

Q = m×C×ΔT

Where Q is heat, m is mass of solution (250g + 14.9g = 264.9g), C is specific heat of solution (It is possible to assume specific heat of pure water, 4.184J/g°C), and ΔT is change in temperature (20.0°C-16.0°C = 4.0°C)

Replacing:

Q = 264.9g×4.184J/g°C×4.0°C

Q = 4433 J = 4.433kJ

As enthalpy is the change in heat per mole of reaction, moles of ammonium chloride that reacted were:

14.9g NH₄Cl × (1mol / 53.491g) = 0.2786 moles

As heat produced per 0.2786moles were 4.433kJ, heat per mole of ammonium chloride is:

4.433kJ / 0.2786mol = 15.91 kJ / mol

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2