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What mass of water is produced by the combustion 1.2 x 10^26 molecules of butane ?

podryga [215]

Answer:

17.934 kg of water

Explanation:

If balanced equation is not given; this format can come in handy.

For any alkane of the type : CₙH₂ₙ₊₂ , it's combustion reaction will follow:

2CₙH₂ₙ₊₂ + (3n+1) O₂ → (2n)CO₂ + 2(n+1) H₂O

For butane:

2C₄H₁₀(g) + 13O₂(g) → 8CO₂(g) + 10H₂O(l)

2 moles of butane gives 10 moles of water.

1 mol of any substance has Avogadro number(N) of molecules in it( 6.022 x 10²³)

Mass of 1 mole of any substance is equal to it's molar mass

So, if 2 x N molecules of butane gives 10 x 18 g of water.

Then 1.2 x 10²⁶ molecules will give:

$\frac{1.2 \times 10^{26} \times 180}{2 \times 6.022 \times 10^{23}}$

= 17.934 x 10³ g of water

= 17.934 kg of water

5 0

3 years ago

What happens to reactants molecules when forming product molecules?

aleksley [76]

Reactant molecules are any substance that goes into a reaction. When forming product molecules, the reactant molecules combine and create a new molecule (i.e. the combination of methane and oxygen can make carbon dioxide and water).

7 0

3 years ago

Two sealed flasks each have a volume of 2.0 liters. Flask A contains N, O gas and Flask B contains H, gas both at 1

barxatty [35]

Answer:

Each gas have same number of molecules.

Explanation:

According to Avogadro law,

Equal volume of all the gases at same temperature and pressure have equal number of molecules.

Mathematical expression:

V ∝ n

V = Kn

V/n = K

k = constant

V = volume of gas

n = number of moles of gas

when volume change is changed from v1 to v2 and number of moles from n1 to n2 this law can be written as,

V1 / n1 = V2 /n2

This state that by increasing the number of moles of gas volume also goes to increase.

4 0

3 years ago

The concentration of protein in a urine sample is calculated to be 2.77 μg/mL. What is the concentration of this solution in uni

maxonik [38]

Answer:

The concentration of this solution in units of pounds per gallon is $2.776*10^{-5} \frac{lb}{gal}$

Explanation:

Units of measurement are established models for measuring different quantities. The conversion of units is the transformation of a quantity, expressed in a certain unit of measure, into an equivalent one, which may or may not be of the same system of units.

In this case, the conversion of units is carried out knowing that 1 μg are equal to 2.205*10⁻⁹ Lb and 1 mL equals 0.00022 Gallons. So

$2.77 \frac{ug}{mL} = \frac{2.77 ug}{mL}$

If 1 μg equals 2.205*10⁻⁹ lb, 2.77 μg how many lb equals?

$lb=\frac{2.77ug*2.205*10^{-9}lb }{1ug}$

lb=6.10785*10⁻⁹

So, 2.77 μg= 6.10785*10⁻⁹ lb

Then:

$2.77 \frac{ug}{mL} = \frac{2.77 ug}{mL}=\frac{6.10785*10^{-9}lb }{mL} =\frac{6.10785*10^{-9}lb }{0.00022 gal} =\frac{6.10785*10^{-9}lb }{0.00022 gal}$

You get:

$2.77 \frac{ug}{mL} = 2.776*10^{-5} \frac{lb}{gal}$

<u><em>The concentration of this solution in units of pounds per gallon is </em></u> $2.776*10^{-5} \frac{lb}{gal}$ <u><em></em></u>

6 0

3 years ago

Calculate how many grams of the first reactant are necessary to completely react with 17.3 g of the second reactant. the reactio

soldier1979 [14.2K]

Taking into account the reaction stoichiometry, 16.611 grams of Na₂CO₃ are necessary to completely react with 17.3 g of CuCl₂.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

Na₂CO₃ + CuCl₂ → CuCO₃ + 2 NaCl

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

Na₂CO₃: 1 mole
CuCl₂: 1 mole
CuCO₃: 1 mole
NaCl: 2 moles

The molar mass of the compounds is:

Na₂CO₃: 129 g/mole
CuCl₂: 134.45 g/mole
CuCO₃: 123.55 g/mole
NaCl: 58.45 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

Na₂CO₃: 1 mole ×129 g/mole= 129 grams
CuCl₂: 1 mole ×134.45 g/mole= 134.45 grams
CuCO₃: 1 mole ×123.55 g/mole= 123.55 grams
NaCl: 2 mole ×58.45 g/mole=116.9 grams

<h3>Mass of CuCl₂ required</h3>

The following rule of three can be applied: If by reaction stoichiometry 134.35 grams of CuCl₂ react with 129 grams of Na₂CO₃, 17.3 grams of CuCl₂ react with how much mass of Na₂CO₃?

mass of Na₂CO₃= (17.3 grams of CuCl₂× 129 grams of Na₂CO₃)÷ 134.35 grams of CuCl₂

<u><em>mass of Na₂CO₃= 16.611 grams</em></u>

Finally, 16.611 grams of Na₂CO₃ is required.

Learn more about the reaction stoichiometry:

brainly.com/question/24741074

brainly.com/question/24653699

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6 0

1 year ago