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Masteriza [31]
3 years ago
13

Chlorine is used by textile manufacturers to bleach cloth. Excess chlorine is destroyed by its reaction with sodium thiosulfate,

Na2S2O3, as follows:
Na2S2O3(aq) + 4Cl2( g) + 5H2O right-arrow 2NaHSO4(aq) + 8HCl(aq)

1)How many moles of HCl can form from 0.21 mol of Cl2?

2)How many moles of H2O are required for the reaction of 0.18 mol of Cl2?

3)How many moles of H2O react if 0.50 mol HCl is formed?
Chemistry
1 answer:
harina [27]3 years ago
6 0
Na2S2O3(aq) + 4Cl2( g) + 5H2O = 2NaHSO4(aq) + 8HCl(aq)

1)How many moles of HCl can form from 0.21 mol of Cl2?
0.21 mol Cl2 ( 8 mol HCl / 4 mol Cl2 ) = 0.42 mol HCl

2)How many moles of H2O are required for the reaction of 0.18 mol of Cl2?
0.18 mol Cl2 ( 5 mol H2O / 4 mol Cl2 ) = 0.225 mol H2O

3)How many moles of H2O react if 0.50 mol HCl is formed?
0.50 mol HCl ( 5 mol H2O / 8 mol HCl ) = 0.3125 mol H2O
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What pressure, in atm, would be exerted by 0.023 grams of oxygen (O2) if it occupies 31.6 mL at 91
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Answer:  A pressure of 0.681 atm would be exerted by 0.023 grams of oxygen (O_2) if it occupies 31.6 mL at 91^{o}C.

Explanation:

Given : Mass of oxygen = 0.023 g

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Convert mL into L as follows.

1 mL = 0.001 L\\31.6 mL = 31.6 mL \times \frac{0.001 L}{1 mL}\\= 0.0316 L

Temperature = 91^{o}C = (91 + 273) K = 364 K

As molar mass of O_2 is 32 g/mol. Hence, the number of moles of O_2 are calculated as follows.

No. of moles = \frac{mass}{molar mass}\\= \frac{0.023 g}{32 g/mol}\\= 0.00072 mol

Using the ideal gas equation calculate the pressure exerted by given gas as follows.

PV = nRT

where,

P = pressure

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n = number of moles

R = gas constant = 0.0821 L atm/mol K

T = temperature

Substitute the value into above formula as follows.

PV = nRT\\P  \times 0.0316 L = 0.00072 mol \times 0.0821 L atm/mol K \times 364 K\\P = \frac{0.00072 mol \times 0.0821 L atm/mol K \times 364 K}{0.0316 L}\\= 0.681 atm

Thus, we can conclude that a pressure of 0.681 atm would be exerted by 0.023 grams of oxygen (O_2) if it occupies 31.6 mL at 91^{o}C.

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