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3 years ago

How many grams of oxygen gas are required to produce 65.75 grams of steam?

Chemistry

1 answer:

Nana76 [90]3 years ago

3 0

Answer:

The mass of oxygen gas required to produce 65.75 grams of steam is approximately 162.2 grams

Explanation:

From the question, we have the following chemical reaction equation;

2C₃H₁₈(l) + 25O₂ (g) → 16CO₂(g) + 18H₂O (g)

The molar mass of oxygen, O₂ = 32 g/mol

The molar mass of steam, H₂O = 18.01528 g/mol

25 moles of oxygen are required to produce 18 moles of steam

Therefore, according to Proust's law of definite proportions;

(32 × 25) g of oxygen are required to produce (18 × 18.01528) g of steam

65.75 g of steam will be produced by (32 × 25)/(18 × 18.01528) × 65.75 g ≈ 162.2 g of oxygen O₂.

You might be interested in

how many grams of potassium chloride, KCL(molar mass is 74.55g/mol) are produced if 25 grams of potassium chlorate KClo^3 decomp

JulsSmile [24]

Answer:

Mass = 15.20 g of KCl

Explanation:

The balance chemical equation for the decomposition of KClO₃ is as follow;

2 KClO₃ = 2 KCl + 3 O₂

Step 1: Calculate moles of KClO₃ as;

Moles = Mass / M/Mass

Moles = 25.0 g / 122.55 g/mol

Moles = 0.204 moles

Step 2: Find moles of KCl as;

According to equation,

2 moles of KClO₃ produces = 2 moles of KCl

So,

0.204 moles of KClO₃ will produce = X moles of KCl

Solving for X,

X = 2 mol × 0.204 mol / 2 mol

X = 0.204 mol of KCl

Step 3: Calculate mass of KCl as,

Mass = Moles × M.Mass

Mass = 0.204 mol × 74.55 g/mol

Mass = 15.20 g of KCl

6 0

3 years ago

Estimate your de Broglie wavelength when you are running. (For this problem use h = 10^−34 in SI units and 1 lb is equivalent to

AlekseyPX

Answer:

A. your running speed 1.5 m/s

B. your mass 70 kg

C. your de Broglie wavelength $6.32x10^{-36}$ m

Explanation:

Hello there!

In this case, since the equation for the calculation of the Broglie wavelength is:

$\lambda =\frac{h}{m*v}$

We can assume a running speed of about 1.5 m/s and a mass of 70 kg, so the resulting Broglie wavelength is:

$\lambda =\frac{6.626x10^{-34}kg\frac{m}{s} }{70kg*1.5\frac{m}{s} }\\\\\lambda =6.32x10^-36m$

Best regards!

7 0

3 years ago

A child has a toy balloon with a volume of 1.80 L. The temperature of the balloon when it was filled was 293 K at a pressure of

Mekhanik [1.2K]

Answer:

2.42L

Explanation:

Given parameters:

V₁ = 1.8L

T₁ = 293K

P₁ = 101.3kPa

P₂ = 67.6kPa

T₂ = 263K

Unknown:

V₂ = ?

Solution:

To solve this problem, we are going to use the combined gas law to find the final volume of the gas. The combined gas law expression combines the equation of Boyle's law, Charles's law and Avogadro's law;

$\frac{P_{1} V_{1} }{T_{1} } = \frac{P_{2} V_{2} }{T_{2} }$

All the units are in the appropriate form. We just substitute and solve for the unknown;

101.3 x 1.8 / 293 = 67.6 x V₂ / 263

V₂ = 2.42L

7 0

3 years ago

At STP, 2.24 L of carbon dioxide gas is collected. What is the number of molecules of gas present in this volume?

Doss [256]

Considering the definition of STP conditions and Avogadro's number, the number of molecules of gas present in 2.24 L is 6.023×10²² molecules.

<h3>Definition of STP condition</h3>

The STP conditions refer to the standard temperature and pressure. Pressure values at 1 atmosphere and temperature at 0 ° C are used and are reference values for gases. And in these conditions 1 mole of any gas occupies an approximate volume of 22.4 liters.

<h3>Definition of Avogadro's number</h3>

Avogadro's Number or Avogadro's Constant is called the number of particles that make up a substance (usually atoms or molecules) and that can be found in the amount of one mole of said substance. Its value is 6.023×10²³ particles per mole. Avogadro's number applies to any substance.

<h3>Amount of molecules of carbon dioxide gas</h3>

At STP, 2.24 L of carbon dioxide gas is collected. So, you can apply the following rule of three: if by definition of STP conditions 22.4 L are occupied by 1 mole of carbon dioxide gas, 2.24 L are occupied by how many moles of carbon dioxide gas?

$amount of moles of carbon dioxide gas=\frac{2.24 Lx1 mole}{22.4 L}$

amount of moles of carbon dioxide gas= 0.1 moles

Finally, 0.1 moles of carbon dioxide gas are collected.

Then you can apply the following rule of three, considering the Avogadro's number: If 1 mole of carbon dioxide gas contains 6.023×10²³ molecules, 0.1 mole of carbon dioxide gas contains how many molecules?

amount of molecules of carbon dioxide gas= (6.023×10²³ molecules × 0.1 moles)÷1 mole

amount of molecules of carbon dioxide gas=6.023×10²² molecules

Finally, the number of molecules of gas present in 2.24 L is 6.023×10²² molecules.

Learn more about

STP conditions:

brainly.com/question/26364483

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Avogadro's Number:

brainly.com/question/11907018

brainly.com/question/1445383

brainly.com/question/1528951