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Tems11 [23]
3 years ago
13

What is the percent yield of O2 if 10.2 g of O2 is produced from the decomposition of 17.0 g of H2O?

Chemistry
2 answers:
yawa3891 [41]3 years ago
5 0

The balanced chemical reaction will be:

2H2O = 2H2 + O2

<span>We are given the amount of water used in the decomposition reaction. This will be our starting point.</span>

<span>17.0 g H2O</span> (1 mol  H2O/ 18.02 g H2O) (1 mol O2/2 mol <span>H2O</span>) ( 32.00 g O2/1mol O2) = 15.09 g O2

Percent yield = actual yield / theoretical yield x 100

<span>Percent yield =10.2 g / 15.09  g x 100</span>

Percent yield = 67.58%

sineoko [7]3 years ago
3 0

Answer: D

Explanation:  67.6

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Propane is often used to heat homes. The combustion of propane follows the following reaction: C3H8(g) + 5O2(g)  3CO2 (g) + 4H
swat32

Answer:

To release 7563 kJ of heat, we need to burn 163.17 grams of propane

Explanation:

<u>Step 1</u>: Data given

C3H8 + 5O2 -----------> 3CO2 + 4H2O      ΔH° = –2044 kJ

This means every mole C3H8

Every mole of C3H8 produces 2044 kJ of heat when it burns (ΔH° is negative because it's an exothermic reaction)

<u>Step 2: </u>Calculate the number of moles to produce 7563 kJ of heat

1 mol = 2044 kJ

x mol = 7563 kJ

x = 7563/2044 =  3.70 moles

To produce 7563 kJ of heat we have to burn 3.70 moles of C3H8

<u>Step 3: </u>Calculate mass of propane

Mass propane = moles * Molar mass

Mass propane = 3.70 moles * 44.1 g/mol

Mass propane = 163.17 grams

To release 7563 kJ of heat, we need to burn 163.17 grams of propane

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Answer:

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Now, using P =nRT/V

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