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Tpy6a [65]
2 years ago
6

I’ll give brainliest please help thanks.

Chemistry
1 answer:
oksano4ka [1.4K]2 years ago
3 0

Answer:

See Explanation

Explanation:

The equation of the reaction is;

C7H16(g) + 11O2(g) ---->7CO2(g) + 8H2O(g)

1a) Number of moles in 228 g of H2O = 228 g/18 g/mol = 12.67 moles

From the reaction equation;

11 moles of O2 yields 8 moles of H2O

x moles of O2 yields 12.67 moles of H2O

x = 11 * 12.67/8

x = 17.4 moles of O2

Since 1 mole of O2 occupies 22.4 L

17.4 moles of O2 occupies 17.4 moles * 22.4 L/1 mole = 389.76 L

1b) Molar mass of C7H16 = 100 g/mol

Number of moles in 300 g of C7H16 = 300 g/100g/mol = 3 moles

1 mole of C7H16 yields 7 moles of CO2

3 moles of C7H16 yields 3 * 7/1 = 21 moles of CO2

If 1 mole of CO2 occupies 22.4 L

21 moles of CO2 occupies 21 moles * 22.4 L/1 mole = 470.4 L of CO2

2) Number of moles in 202 L of H2 is obtained by;

1 mole of H2 occupies 22.4 L

x moles of  H2 occupies 202 L

x = 1 mole * 202 L/22.4 L =  9 moles of H2

From the reaction equation;

3 moles of H2 yields 2 moles of PH3

9 moles of H2 will yield 9 * 2/3 = 6 moles of PH3

Mass of 6 moles of PH3 = 6 moles of PH3 * 34 g/mol = 204 g of PH3

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4 0
1 year ago
Submit your answer for the remaining reagent in Tutorial Assignment #1 Question 9 here, including units. Note: Use e for scienti
djverab [1.8K]
<h3>Answer:</h3>

The mass of excessive water (H₂O) is 40.815 kg

<h3>Explanation:</h3>

The Equation for the reaction is;

Li₂O(s) + H₂O(l) → 2LiOH(s)

From the question;

Mass of water removed is 80.0 kg

Mass of available Li₂O is 65.0 kg

We are required to calculate the mass of excessive reagent.

<h3>Step 1: Calculating the number of moles of water to be removed</h3>

Moles = Mass ÷ Molar mass

Molar mass of water = 18.02 g/mol

Mass of water = 80 kg (but 1000 g = 1kg)

                        = 80,000 g

Therefore;

Moles of water = \frac{80,000g}{18.02 g/mol}

                = 4.44 × 10³ moles

<h3>Step 2: Moles of Li₂O available </h3>

Moles = mass ÷ molar mass

Mass of Li₂O  available = 65.0 kg or 65,000 g

Molar mass Li₂O  = 29.88 g/mol

Moles of Li₂O  = 65,000 g ÷ 29.88 g/mol

          = 2.175 × 10³ moles Li₂O

<h3>Step 3: Mass of excess reagent </h3>

From the equation; Li₂O(s) + H₂O(l) → 2LiOH(s)

1 mole of Li₂O reacts with 1 mole of water to form two moles of LiOH

The ratio of Li₂O to H₂O is 1:1

  • Thus, 2.175 × 10³ moles of Li₂O will react with 2.175 × 10³ moles of water.
  • However, the number of moles of water to be removed is 4.44 × 10³ moles  but only 2.175 × 10³ moles will react with the available Li₂O.
  • This means, Li₂O  is the limiting reactant while water is the excessive reagent.

Therefore:

Moles of excessive water =  4.44 × 10³ moles  - 2.175 × 10³ moles

                                           = 2.265 × 10³ moles

Mass of excessive water = 2.265 × 10³ moles × 18.02 g/mol

                                          = 4.0815 × 10⁴ g or

                                          = 40.815 kg

Thus, the mass of excessive water is 40.815 kg

7 0
2 years ago
1.) What coefficient values will balance the reaction shown? CH4 + O2 ?CO2 + H2O
Varvara68 [4.7K]

Answer:

It's simpler than most questions :)

Explanation:

1) B

2) B

3) D

4) A

5) A

Hope this will help :)

5 0
3 years ago
Earth’s inner core:_____ as Earth’s outer core:liquid
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Answer:However, unlike the outer core, the inner core is not liquid or even molten. The inner core's intense pressure—the entire rest of the planet and its atmosphere—prevents the iron from melting. The pressure and density are simply too great for the iron atoms to move into a liquid state.

Explanation:please give brainliest

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1 year ago
Beryllium oxide, Beo, is an electrical insulator. How
egoroff_w [7]

Answer:

There are 10.0 moles of beryllium oxide in a 250 grams sample of the compound.

Explanation:

We can calculate the number of moles (η) of BeO as follows:

\eta = \frac{m}{M}

Where:

m: is the mass = 250 g

M: is the molar mass = 25.0116 g/mol

Hence, the number of moles is:

\eta = \frac{250 g}{25.0116 g/mol} = 10.0 moles

Therefore, there are 10.0 moles of beryllium oxide in a 250 grams sample of the compound.  

I hope it helps you!

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3 years ago
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