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Sveta_85 [38]
3 years ago
8

In lab you reacted 60 grams of propane (C3H8) with 13 grams of oxygen gas (O2). Your

Chemistry
1 answer:
Andre45 [30]3 years ago
5 0

Answer:

120g

Explanation:

Based on the reaction, when 1 mole of propane and 5 moles of oxygen react, 3 moles of CO2 and 4 moles of H2O are produced. The ratio of production is 3 moles of CO2 per 4 moles of H2O.

Thus, we need to convert mass of water to moles using its molar mass:

Moles H2O (Molar mass: 18g/mol):

63g H2O * (1mol / 18g) = 3.5 moles H2O

Converting to moles of CO2:

3.5 moles H2O * (3 moles CO2 / 4 moles H2O) = 2.625 moles CO2

Mass CO2 (Molar mass: 44g/mol):

2.625 moles CO2 * (44g / mol) = 115.5g of CO2 are released

<h3>120g </h3>

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If 5.0 liters H2 (g) at STP is heated to a temperature of 985, pressure remaining constant, the new volume of the gas will be?
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Answer:

V_2=18 \ L \ H_2

General Formulas and Concepts:

<u>Chemistry - Gas Laws</u>

  • STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K
  • Charles' Law: \frac{V_1}{T_1} =\frac{V_2}{T_2}

Explanation:

<u>Step 1: Define</u>

Initial Volume: 5.0 L H₂ gas

Initial Temp: 273 K

Final Temp: 985 K

Final Volume: ?

<u>Step 2: Solve for new volume</u>

  1. Substitute:                    \frac{5 \ L \ H_2}{273 \ K} =\frac{x \ L \ H_2}{985 \ K}
  2. Cross-multiply:             (5 \ L \ H_2)(985 \ K) = (x \ L \ H_2)(273 \ K)
  3. Multiply:                        4925 \ L \ H_2 \cdot K = 273x \ L \ H_2 \cdot K
  4. Isolate <em>x</em>:                       18.0403 \ L \ H_2 = x
  5. Rewrite:                         x=18.0403 \ L \ H_2

<u>Step 3: Check</u>

<em>We are given 2 sig figs as the smallest. Follow sig fig rules and round.</em>

<em />18.0403 \ L \ H_2 \approx 18 \ L \ H_2<em />

5 0
3 years ago
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