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lozanna [386]
3 years ago
13

Lana is studying a chemical reaction in which the reactants propane (c3h8) and oxygen gas (o2) react to yield the products carbo

n dioxide (co2) and water (h2o). lana's teacher informs her that two molecules of propane and ten molecules of oxygen are reacting. the teacher then asks lana to assess how many molecules of carbon dioxide and water will be produced in the reaction. which answer choice correctly describes what lana should do to determine the answer to her teacher's question?
Chemistry
1 answer:
makvit [3.9K]3 years ago
8 0
In such type of problems where reactant and product ratios are calculated one (Launa) should take start from balanced chemical equation. A balanced chemical equations helps in solving such problems. For this purpose chemical equation of this combustion reaction is drawn.

Equation:

                          C₃H₁₀ ⁺ 5 O₂  ⁻⁻⁻⁻⁻⁻⁻⁻⁻>  3 CO₂ ⁺ 4 H₂O

Now, According to this balanced equation one molecule of propane reacts with 5 molecules of oxygen to produce 3 molecules of carbon dioxide and 4 molecules of water.

As She is given with 2 molecules of propane and 10 molecules of oxygen, which means that the equation is multiplied by 2. So, the product side must also be multiplied by 2 to balance the equatio (Law of Conservation of Mass).

Result:

                      ( C₃H₁₀ ⁺ 5 O₂  ⁻⁻⁻⁻⁻⁻⁻⁻⁻>  3 CO₂ ⁺ 4 H₂O ) ₓ 2

                        2 C₃H₁₀ ⁺ 10 O₂  ⁻⁻⁻⁻⁻⁻⁻⁻⁻>  6 CO₂ ⁺ 8 H₂O

So, 6 molecules of carbon dioxide and 8 molecules of water are produced.
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Formula units in 450 g of Na_{2} So_{4} is 1.93 × 10²⁴ formula units.

<u>Explanation:</u>

First we have to find the number of moles in the given mass by dividing the mass by its molar mass as,

$\frac{450 g}{142.04 g/mol} =  3.2 moles

Now, we have to multiply the number of moles of Na₂SO₄ by the Avogadro's number, 6.022 × 10²³ formula units/mol, so we will get the number of formula units present in the given mass of the compound.

3.2 mol × 6.022 × 10²³ = 1.93 × 10²⁴ formula units.

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The molality of the solution = 17.93 m

<h3>Further explanation</h3>

Given

6.00 L water with 6.00 L of  ethylene glycol(ρ=1.1132 g/cm³= 1.1132 kg/L)

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mol of solute = mol of ethylene glycol

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3.07 × 10^4=30,700

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