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Volgvan
3 years ago
11

An aqueous solution is analyzed and found to contain potassium ions and sulfite ions. write the equation for the dissolution of

the salt that produced this solution. if the solution contains 0.700 mol of potassium ions, how many moles of sulfite ions are present? answer in units of ol.
Chemistry
1 answer:
enot [183]3 years ago
7 0
1) T<span>he dissolution of the salt potassium sulfite:
K</span>₂SO₃(aq) → 2K⁺(aq) + SO₃²⁻(aq).
Potassium has +1 charge because it lost one electron to accomplish stabile electron configuration of noble gas argon.
2) From dissolution reaction: n(K⁺) : n(SO₃²⁻) = 2 : 1.
n(K⁺) = 0.700 mol.
0.700 mol : n(SO₃²⁻) = 2 : 1.
n(SO₃²⁻) = 0.700 mol ÷ 2.
n(SO₃²⁻) = 0.350 mol; amount of sulfite anions.
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3 years ago
Part 1. A chemist reacted 18.0 liters of F2 gas with NaCl in the laboratory to form Cl2 gas and NaF. Use the ideal gas law equat
Alika [10]

Answer:

Part 1

The mass of the NaCl that reacted with F₂ at 290.K and 1.5 atm is approximately 132.6 gams

Part 2

The mass of NaCl that can react with the same volume of gas at STP is approximately 93.77 grams

Explanation:

Part 1

The volume of F₂ gas in the reaction, V = 18.0 liters

The ideal gas equation is P·V = n·R·T

∴ n = P·V/(R·T)

The pressure, P = 1.5 atm

The temperature, T = 290 K

The universal gas constant, R = 0.0820573 L·atm/(mol·K)

∴ n = 1.5×18/(0.0820573 × 290) ≈ 1.134615

The number of moles of F₂ in the reaction n ≈ 1.134615 moles

The chemical reaction is given as follows;

F₂ + 2NaCl → Cl₂ + 2NaF

1 mole of F₂ reacts with 2 moles of NaCl

Therefore;

1.134615 moles of F₂ reacted with 2 × 1.134615 moles ≈ 2.26923 moles of NaCl

1 mole of NaCl = The molar mass of NaCl, MM = 58.44 g/mol

The mass, of 2.26923 moles of NaCl, m = Number of moles × MM

∴ m ≈ 2.26923 moles × 58.44 g/mol ≈ 132.6 grams

The mass of the NaCl ≈ 132.6 gams

Part 2

The volume occupied by 1 mole of all gases at STP = 22.4 l/mole

Therefore, the number of moles of F₂ in 18.0 L of F₂ = 18.0 L/(22.4 L/mole) ≈ 0.804 moles

Therefore;

The number of moles of NaCl, in the reaction n = 2 × The number of moles of F₂ ≈ 2×0.804 moles = 1.608 moles

The number of moles of NaCl, in the reaction n ≈ 1.608 moles

The mass of NaCl in the reaction, m = n × MM

∴ m ≈ 1.608 moles × 58.44 g/mol ≈ 93.97 grams

The mass of NaCl that can react with the same volume of gas at STP ≈ 93.77 grams

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Answer:

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