Question

would you please review my answers and help me with the ones that not answered because im stuck. questions 1-9

the mass of water us 3.549g

the initial temp of water is 12.19 degree C

the final temperature of the water is -14.01 degree C

How much of each reactant do you need to rernove enough heat to freeze the water? That is the question you are going to answer. A The system, surroundings and the flow of beat In this change, we will consider the solid reactants in the beaker as the system: Ba(OH). (s) + 2NHCIO → BaCl, (s) + 2 NH, (g) + 2 H20 (1) AH° (kJ/mol) -626.56 -314.55 -495.73 45.90 -285.83 and the water on the block as the surroundings. The reaction above occurs with an increase in enthalpy, so it pulls" in heat from the surroundings (the water). The heat absorbed in the reaction in (1) can be calculated by using AH, to find AHU However, there are three changes that can take place as heat leaves the water. H0 (room temperature) →H,0 (1,0°C) (2) HO (1, 0 °C) HEO (s, 0°C) (3) HO (5, 0 °C) →H0 (s, final temperature) (4) These three changes can be shown as follows:

Answer 1

Mass of water = 3.549 g

Initial water temperature= 12.19°C

Final temperature = -14. 01°C

There are three stages of heat transfer

Heat extracted from 12.19°C to 0°C

Latent heat at 0°C

Heat extracted from 0°C to -14. 01 °C

1) cp of water = 1cal/g

Q1 = -3.549(1) (12.19-0) = -43.26 cal

2) latent heat of fusion of water = 80 cal/g

Q2 =- 3.549(80) = -283. 92 cal

3)

Cp of ice = 0.5 cal

Q3 = -3.549(0.5)(14.01) -24.86 cal

4) Total heat lost to surrounding =

Q = -24. 86-283.92-43.26= -352. 04 cal

5) heat absorbed by system

No heat losses, hence heat lost to surroundings = heat absorbed by system

Q = +352.04 cal

6) The reaction occuring is

$Ba(OH)_{2}+2NH_{4}Cl\rightarrow BaCl_{2}+2NH_{3}+2H_{2}O$

Hf(Barium hydroxide) -626. 56 KJ/mol

Hf(ammonium chloride) =-314. 55 KJ/mol

Hf(barium chloride) = -495. 73 KJ/mol

Hf(ammonia) = -45. 90 KJ/mol

Hf(water) = -285. 83 KJ/mol

∆Hr = nHproducts - nHreactants

Hproducts= 1(-495.73) +(2) (-45.80) +(2) (-285.83) = -1158.99 KJ/mol

Hreactant = (2)(-314.55) +(1) (-626.56) =

- 1255.66 KJ/mol

∆Hr = -1158.99-(-1255.66 )= + 96.67 KJ/mol reactant

The reaction is endothermic and it absorbs heat

7)

Heat absorbed by system = heat lost by water = +352.04 cal

1 cal = 4.184 J

Q = 352.04(4.184)= +1472.95 J = 1.472 KJ

Heat produced by reacting 1 mole barium hydroxide = ∆Hr = 96.67 KJ

Number of moles of barium hydroxide that absorbs 1.47295 KJ heat =( 1.472/96.67) = 0.01523 moles

8)

Moles of barium hydroxide reacted =

0.01523 moles

According to stiochiometry , moles of ammonium chloride required = (2) (0.01523) = 0.03046 moles

9) M. W of Ba(OH) 2 = 137.32+(32) +(2) =

171.32 g/mol

M. W of NH4Cl = (14+4+35.5) = 53.5 g/mol

Grams of each reactant required =

Ba(OH) 2 = 0.01523(171.32) = 2.609 g

NH4Cl = 0.03046(53.5) = 1.629 g

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