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Sever21 [200]
3 years ago
9

If 5.738 grams of AgNO3 is mixed with 4.115 grams of BaCl2 and allowed to react according to the balanced equation: BaCl2(aq) +

2 AgNO3(aq) → 2 AgCl(s) + Ba(NO3)2(aq) What is the limiting reagent? BaCl2AgNO3 How many grams of AgCl could be produced? grams AgCl What mass, in grams, of the excess reagent will remain? grams of excess reagent
Chemistry
1 answer:
IceJOKER [234]3 years ago
5 0

Answer:

Limiting reagent: AgNO3

grams AgCl : 2.44 g AgCl

grams of excess reagent remain: 0.62 g BaCl2

Explanation:

1. Change grams to mol:

<em>AgNO3:</em>

5.738g x (1mol/169.87g) = 0.034 mol AgNO3

<em>BaCl2:</em>

4.115g x (1 mol/208.23g) = 0.020 mol BaCl2

2. Limiting reagent:

<em>AgNO3:</em>

0.034 mol AgNO3 x (1 mol BaCL2/ 2mol AgNO3) = 0.017 mol BaCl2

<em>BaCl2:</em>

0.020 mol BaCl2 x (2 mol AgNO3/1 mol BaCl2) = 0.04 mol AgNO3

Limiting reagent: AgNO3

3. Grams of AgCl produced:

Using the limiting reagent:

0.017 mol AgNO3 x (2mol AgCl / 2 mol AgNO3) = 0.017 mol AgCl

4. Change mol to grams:

0.017 mol AgCl x ( 143.32 g AgCl /1mol AgCl) =2.44 g AgCl

5. Grams of the excess reagent:

0.034 mol AgNO3 x (1 mol BaCl2 / 2 mol AgNO3) = 0.017 mol BaCl2

0.020 mol BaCl2 - 0.017 mol BaCl2 = 0.003 mol BaCl2

0.003 mol BaCl2 x ( 208.23 g BaCl2 / 1 mol BaCl2) = 0.62 g BaCl2

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b. 939.43 cm^{3}

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d. 64.5J

Explanation:

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Thus, n = PV/RT = (125000 × 0.000785)/(8.314 × 400) = 0.03 mol

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To calculate this, we use the constant pressure process;

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Where q is 83.8J according to the question

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83.8 = (0.03 × 35.5) (T_{1} - 400K)

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To calculate this, we make use of the Charles' law(Temperature and pressure are directly proportional)

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V_{1}   = (785 × 478.69)/400

V_{1}   = 939.43 cm^{3}

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Mathematically, the work done by the system is calculated as follows;

w = P(V_{1}- V_{o}) = 125 KPa ( 939.43 - 785) = 19.30 J

d. Change in internal energy of the steam in J

ΔU = q - w = 83.8 - 19.3 = 64.5J

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