The combustion of a sample of butane, C4H10 (lighter fluid), produced 2.46 grams of water.
2 C4H10 + 13O2 -------> 8CO2 + 10H2O
(a) How many moles of water formed?
(b) How many moles of butane burned?
(c) How many grams of butane burned?
(d) How much oxygen was used up in moles?
(e) How much oxygen was used up in grams?

Answers

Answer 1

a. 0.137

b. 0.0274

c. 1.5892 g

d. 0.1781

e. 5.6992 g

Further explanation

Given

Reaction

2 C4H10 + 13O2 -------> 8CO2 + 10H2O

2.46 g of water

Required

moles and mass

Solution

a. moles of water :

2.46 g : 18 g/mol = 0.137

b. moles of butane :

= 2/10 x mol water

= 2/10 x 0.137

= 0.0274

c. mass of butane :

= 0.0274 x 58 g/mol

= 1.5892 g

d. moles of oxygen :

= 13/2 x mol butane

= 13/2 x 0.0274

= 0.1781

e. mass of oxygen :

= 0.1781 x 32 g/mol

= 5.6992 g

Answer 2

0.137moles of water is formed, 0.0274 moles or 1.5892 g of butane burned and 0.1781 moles or 5.6992 g of O₂ is used.

What is combustion reaction?

Those reaction in which fuel is oxidized by the oxygen molecules and produce carbon dioxide and water molecule.

Given chemical reaction is:
2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O

Given mass of water = 2.46 grams

Moles will be calculated as:

n = W/M, where

W = given mass

M = molar mass

(a) Moles of water formed is calculated as:

Moles of water n = 2.46g / 18 g/mol = 0.137moles

(b) From the stoichiometry of the reaction, it is clear that:

10 moles of water = produced by 2 moles of butane

0.137 moles of water = produced by 2/10×0.137=0.0274 moles of butane

(c) Weight of butane is calculated by using moles:

W = 0.0274 × 58 g/mol = 1.5892 g

(d) From the stoichiometry of the reaction, it is clear that:

2 moles of butane = react with 13 moles of O₂

0.027 moles of butane = react with 13/2×0.027=0.1781 moles of O₂

(e) Mass of oxygen is calculated as:
W = 0.1781 x 32 g/mol = 5.6992 g

Hence, (a) 0.137moles, (b) 0.0274 moles, (c) 1.5892 g, (d) 0.1781 moles and (e) 5.6992 g.

To know more about combustion reaction, visit the below link:

https://brainly.com/question/9425444


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Answer:

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Answers

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Further explanation

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The average speed

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Answers

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Answers

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Answers

Answer:

See explanation

Explanation:

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1. Cesium (Cs), Rubidium (Rb), and Sodium (Na)

2. Barium (Ba), Calcium (Ca), and Beryllium (Be)

3. Fluorine (F), Chlorine (Cl), and Bromine (Br)

4. Tellurium (Te), Oxygen (O), and Sulfur (S)

Answers

1. The reactivity among the alkali metals increases as you go down the group due to the decrease in the effective nuclear charge from the increased shielding by the greater number of electrons. The greater the atomic number, the weaker the hold on the valence electron the nucleus has, and the more easily the element can lose the electron. Conversely, the lower the atomic number, the greater pull the nucleus has on the valence electron, and the less readily would the element be able to lose the electron (relatively speaking). Thus, in the first set comprising group I elements, sodium (Na) would be the least likely to lose its valence electron (and, for that matter, its core electrons).

2. The elements in this set are the group II alkaline earth metals, and they follow the same trend as the alkali metals. Of the elements here, beryllium (Be) would have the highest effective nuclear charge, and so it would be the least likely to lose its valence electrons. In fact, beryllium has a tendency not to lose (or gain) electrons, i.e., ionize, at all; it is unique among its congeners in that it tends to form covalent bonds.

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4. The pattern for the elements in this set (the group VI chalcogens) generally follows that of the halogens. The greater the atomic number, the weaker the pull of the nucleus, and so the lesser the tendency to gain electrons. Tellurium (Te) has the highest atomic number among the elements in the set, and so it would be the least likely to gain electrons.

The major product during addition of HBr to propene in the presence of peroxide is
A. 2-Bromopropane
B. 1-Bromopropane
C. 1,1-dibromopropane
D. 1,2-Dibromopropane

Answers

Answer:

A. 2-Bromopropane

Explanation:

Propene is the second member of the alkene group with the chemical formula; C3H6. It reacts with hydrogen bromide (HBr), which is an acidic compound to form an alkylbromide.

In this reaction, the formation of a major product is in accordance with Markovnikov's rule, which emphasizes that hydrogen will be added to the carbon with more hydrogen. Hence, the bromide ion of HBr will be added to the second carbon atom of the alkene (propene), to form 2-bromopropane as a major product.

The reaction is as follows:

CH3CH=CH2 (propene) + HBr → CH3CHBrCH3 (2-bromopropane).

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Answers

Answer:

Explanation:

The Molar mass of Fe(Bro)2 = 167. 748 mol/g

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