A sample of an ideal gas at 1.00 atm and a volume of 1.50 L was placed in a weighted balloon and dropped into the ocean. As the
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The percent yield of carbon dioxide will be 49.0 %.
<h3>Percent yield</h3>First, let's look at the equation of the reaction:
The mole ratio of octane to oxygen is 2:25.
Mole of 3.43 g octane = 3.43/114.23 = 0.03 mol
Mole of 19.1 g oxygen = 19.1/32 = 0.60 mol
Thus, octane is limiting.
Mole ratio of octane to carbon dioxide = 2:16.
Equivalent mole of carbon dioxide = 0.03 x 8 = 0.24 mol
Mass of 0.24 mol carbon dioxide = 0.24 x 44.01 = 10.5624 grams
Percent yield of carbon dioxide = 5.18/10.5624 = 49.0 %
More on percent yield can be found here: brainly.com/question/17042787
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1.386 g of Mg ribbon combusts to form 2.309 g of oxide product. The mass percent of oxygen in the oxide is 40.0 %.
Let's consider the reaction for the combustion of Mg.
Mg + 1/2 O₂ ⇒ MgO
1.386 g of Mg combusts to form 2.309 g of MgO. We want to determine the mass of oxygen in MgO. According to Lavoisier's law of conservation of mass, matter is not created nor destroyed over the course of a chemical reaction. Then, the mass of Mg in the reactants is equal to the mass of Mg in MgO. The mass of the magnesium oxide is the sum of the masses of magnesium and oxygen. The <u>mass of oxygen in the oxide</u> is:
We can calculate the mass percent of O in MgO using the following expression.
You can learn more about mass percent here: brainly.com/question/14990953