What mass of KO2, in grams, is required to react with 15.9 L of CO2 at 23.0°C and 955 mm Hg? _ g (b)(c)

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(a)What mass of KO2, in grams, is required to react with 15.9 L of CO2 at 23.0°C and 955 mm Hg? _ g (b)(c)

A self-contained underwater breathing apparatus uses canisters containing potassium superoxide. The superoxide consumes the exhaled CO2 by a person and replaces it with oxygen.

4KO2 (s) + 2CO2 (g) —> 2K2CO3 (s) + 3 O2 (g)

Hydrazine reacts with O2 according to the following equation: N2H4 (g) + O2 (g) —> N2 (g) + 2H2O (l)

Assume the O2 needed for the reaction is in a 650. L tank at 16.0 °C. What must the oxygen pressure be in the tank to have enough oxygen to consume 1.28 kg of hydrazine completely? ____atm

A 0.0125-g sample of a gas with an empirical formula of CHF2 is placed in a 165-mL flask. It has a pressure of 6.85 mm Hg at 22.5 °C. What is the molecular formula of the compound?

The Correct Answer and Explanation is:

Let’s go through each part of the question in sequence.

(a) What mass of KO₂ is required to react with 15.9 L of CO₂ at 23.0 °C and 955 mm Hg?

Given reaction: 4KO2(s)+2CO2(g)→2K2CO3(s)+3O2(g)4KO_2 (s) + 2CO_2 (g) \rightarrow 2K_2CO_3 (s) + 3O_2 (g)

We first calculate the number of moles of CO₂ using the ideal gas law: PV=nRTPV = nRT

Convert units first:

  • P=955 mmHg=955760 atm≈1.2566 atmP = 955\ \text{mmHg} = \frac{955}{760}\ \text{atm} \approx 1.2566\ \text{atm}
  • V=15.9 LV = 15.9\ \text{L}
  • T=23.0 ∘C=296.15 KT = 23.0\ ^\circ C = 296.15\ \text{K}
  • R=0.0821 L\cdotpatm/mol\cdotpKR = 0.0821\ \text{L·atm/mol·K}

n=PVRT=(1.2566)(15.9)(0.0821)(296.15)≈0.8215 mol CO₂n = \frac{PV}{RT} = \frac{(1.2566)(15.9)}{(0.0821)(296.15)} \approx 0.8215\ \text{mol CO₂}

From the reaction:
2 mol CO₂ → 4 mol KO₂ → mol ratio = 2:4 = 1:2 mol KO₂=0.8215×2=1.643 mol\text{mol KO₂} = 0.8215 \times 2 = 1.643\ \text{mol}

Molar mass of KO₂ = 39.10 (K) + 2×16.00 (O) = 71.10 g/mol Mass KO₂=1.643×71.10≈116.8 g\text{Mass KO₂} = 1.643 \times 71.10 \approx \boxed{116.8\ \text{g}}

(b) What must the oxygen pressure be in the tank to consume 1.28 kg of hydrazine?

Reaction: N₂H₄ (g) + O₂ (g) → N₂ (g) + 2H₂O (l)\text{N₂H₄ (g) + O₂ (g) → N₂ (g) + 2H₂O (l)}

Step 1: Moles of N₂H₄:

Molar mass N₂H₄ = 2×14.01 + 4×1.008 = 32.05 g/mol Mass=1.28 kg=1280 g, mol=128032.05≈39.93 mol\text{Mass} = 1.28\ \text{kg} = 1280\ \text{g},\ \text{mol} = \frac{1280}{32.05} \approx 39.93\ \text{mol}

From the equation: 1 mol N₂H₄ reacts with 1 mol O₂ → mol O₂ = 39.93 mol

Step 2: Use PV = nRT to find pressure.

  • V=650.0 LV = 650.0\ \text{L}
  • T=16.0°C=289.15 KT = 16.0°C = 289.15\ \text{K}
  • R=0.0821R = 0.0821

P=nRTV=(39.93)(0.0821)(289.15)650.0≈1.46 atmP = \frac{nRT}{V} = \frac{(39.93)(0.0821)(289.15)}{650.0} \approx \boxed{1.46\ \text{atm}}

(c) Determine the molecular formula of a gas with empirical formula CHF₂

Given:

  • Mass = 0.0125 g
  • Volume = 165 mL = 0.165 L
  • Pressure = 6.85 mmHg = 6.85 / 760 = 0.00901 atm
  • Temperature = 22.5°C = 295.65 K
  • R = 0.0821

Using PV = nRT: n=PVRT=(0.00901)(0.165)(0.0821)(295.65)≈6.13×10−6 moln = \frac{PV}{RT} = \frac{(0.00901)(0.165)}{(0.0821)(295.65)} \approx 6.13 \times 10^{-6}\ \text{mol}

Now calculate molar mass: M=massmol=0.01256.13×10−6≈204 g/molM = \frac{\text{mass}}{\text{mol}} = \frac{0.0125}{6.13 \times 10^{-6}} \approx 204\ \text{g/mol}

Empirical formula = CHF₂
Empirical mass = 12.01 (C) + 1.008 (H) + 2×18.998 (F) ≈ 52.01 g/mol Ratio=20452.01≈3.92≈4\text{Ratio} = \frac{204}{52.01} \approx 3.92 \approx 4

Molecular formula = (CHF2)4=C4H4F8(CHF₂)_4 = \boxed{C₄H₄F₈}

✅ Final Answers:

(a) 116.8 g KO₂
(b) 1.46 atm O₂
(c) Molecular formula: C₄H₄F₈

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