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tigry1 [53]
3 years ago
15

A 1.52 g sample of KCIO3 is reacted according to the balanced equation below. How many liters of O2 is produced at a pressure of

0.974 atm and a
temperature of 37'C?
2 KCIO3(s) ---> 2 KCI (s) + 3 O2(9)
Chemistry
1 answer:
ipn [44]3 years ago
7 0

Answer:

0.486 L

Explanation:

Step 1: Write the balanced reaction

2 KCIO₃(s) ⇒ 2 KCI (s) + 3 O₂(g)

Step 2: Calculate the moles corresponding to 1.52 g of KCIO₃

The molar mass of KCIO₃ is 122.55 g/mol.

1.52 g × 1 mol/122.55 g = 0.0124 mol

Step 3: Calculate the moles of O₂ produced from 0.0124 moles of KCIO₃

The molar ratio of KCIO₃ to O₂ is 2:3. The moles of O₂ produced are 3/2 × 0.0124 mol = 0.0186 mol

Step 4: Calculate the volume corresponding to 0.0186 moles of O₂

0.0186 moles of O₂ are at 37 °C (310 K) and 0.974 atm. We can calculate the volume of oxygen using the ideal gas equation.

P × V = n × R × T

V = n × R × T/P

V = 0.0186 mol × (0.0821 atm.L/mol.K) × 310 K/0.974 atm = 0.486 L

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Answer:

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Answer:

Net ionic equation: Ag⁺(aq) + Cl⁻(aq) → AgCl(s)

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Explanation:

When sodium nitrate, NaNO₃ and silver nitrate, AgNO₃ are dissolved in water, the Na⁺, NO₃⁻ and Ag⁺ ions are formed.

Then, the addition of NaCl (Na⁺ and Cl⁻) produce AgCl⁻ as precipitate. <em>The net ionic equation is:</em>

<h3>Ag⁺(aq) + Cl⁻(aq) → AgCl(s)</h3><h3 />

If 2.54g of AgCl are formed and represents the 95.9% of yield. The real amount of AgCl is:

2.54g AgCl * (100% / 95.9%) = 2.65g AgCl.

In moles (Molar mass AgCl = 143.32g/mol):

2.65g AgCl * (1mol / 143.32g) = 0.0185 moles of AgCl = Moles of AgNO₃

<em>Because all Ag comes from AgNO₃</em>

<em />

Thus, the original mass of silver nitrate and its precentage is (Molar mass AgNO₃ = 169.87g/mol):

0.0185 moles AgNO₃ * (169.87g / mol) = 3.14g of AgNO₃

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The combustion of 43.9 g of ammonia with 258 g of oxygen produces ________ g of no2.
adell [148]
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Answer:

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