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JulijaS [17]
3 years ago
5

Coal power plants burn large amounts of coal, c(s), in an o 2 ​ (g) atmosphere to generate electricity. the chemical reaction re

sponsible for producing this energy is shown below: c(s) + o 2 ​ (g) → co 2 ​ (g) determine the volume of co 2 ​ in liters produced when 100 metric ton of c(s) is completely burned in an o 2 ​ atmosphere. the density of co 2 ​ is 1.98 kg/m 3 (1 metric ton = 1000 kg; 1 m 3 = 1000 l)
Chemistry
1 answer:
charle [14.2K]3 years ago
6 0
Answer is: volume of carbon dioxide is 1,84·10⁸ l.
Chemical reaction: C + O₂ → CO₂.
m(C) = 100 t · 1000 kg/t = 100000 kg 
m(C) = 100000 kg · 1000 g/kg = 10⁸ g.
n(C) = m(C) ÷ M(C).
n(C) = 10⁸ g ÷ 12 g/mol.
n(C) = 8,33·10⁶ mol.
From chemical reaction: n(C) . n(CO₂) = 1 : 1.
n(CO₂) = 8,33·10⁶ mol.
m(CO₂) = 8,33·10⁶ mol · 44 g/mol.
m(CO₂)  = 3,66·10⁸ = 3,66·10⁵ kg.
V(CO₂) = 3,66·10⁵ kg ÷ 1,98 kg/m³ = 1,84·10⁵ m³.
V(CO₂) = 1,84·10⁵ m³ · 1000 l/m³ = 1,84·10⁸ l.

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onsider the following reaction: CaCN2 + 3 H2O → CaCO3 + 2 NH3 105.0 g CaCN2 and 78.0 g H2O are reacted. Assuming 100% efficiency
mestny [16]

Answer : The excess reactant is, H_2O

The leftover amount of excess reagent is, 7.2 grams.

Solution : Given,

Mass of CaCN_2 = 105.0 g

Mass of H_2O = 78.0 g

Molar mass of CaCN_2 = 80.11 g/mole

Molar mass of H_2O = 18 g/mole

Molar mass of CaCO_3 = 100.09 g/mole

First we have to calculate the moles of CaCN_2 and H_2O.

\text{ Moles of }CaCN_2=\frac{\text{ Mass of }CaCN_2}{\text{ Molar mass of }CaCN_2}=\frac{105.0g}{80.11g/mole}=1.31moles

\text{ Moles of }H_2O=\frac{\text{ Mass of }H_2O}{\text{ Molar mass of }H_2O}=\frac{78.0g}{18g/mole}=4.33moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

CaCN_2+3H_2O\rightarrow CaCO_3+2NH_3

From the balanced reaction we conclude that

As, 1 mole of CaCN_2 react with 3 mole of H_2O

So, 1.31 moles of CaCN_2 react with 1.31\times 3=3.93 moles of H_2O

From this we conclude that, H_2O is an excess reagent because the given moles are greater than the required moles and CaCN_2 is a limiting reagent and it limits the formation of product.

Left moles of excess reactant = 4.33 - 3.93 = 0.4 moles

Now we have to calculate the mass of excess reactant.

\text{ Mass of excess reactant}=\text{ Moles of excess reactant}\times \text{ Molar mass of excess reactant}(H_2O)

\text{ Mass of excess reactant}=(0.4moles)\times (18g/mole)=7.2g

Thus, the leftover amount of excess reagent is, 7.2 grams.

8 0
3 years ago
Which dilute acid is used to make sodium carbonate salt ​
andreev551 [17]
Hydrochloric acid :))
8 0
3 years ago
Read 2 more answers
Nitration of 1,4−dimethylbenzene (p−xylene) gives a single product having the molecular formula c8h9no2 in high yield. draw this
GarryVolchara [31]
Structure is in document below.
The mononitration of p-xylene can be easily carried out at 30 degrees C.
Para-xylene<span> (</span><span>p-xylene</span><span>) is an </span>aromatic hydrocarbon, <span>one of the three </span>isomers<span> of </span>dimethylbenzene. Para-xylene is colorless and highly flammable, not acutely toxic and has some <span>narcotic effects.</span>
Download docx
8 0
3 years ago
Which two types of information are written in an element's box in the periodic table?
GenaCL600 [577]

Answer:

Yes it is B,D.

Explanation:

Each box represents an element and contains its atomic number, symbol, average atomic mass, and (sometimes) name. The elements are arranged in seven horizontal rows, called periods or series, and 18 vertical columns, called groups.

8 0
2 years ago
Which of the following equations are balanced?
Debora [2.8K]

Answer:

The only balanced equation is:

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Explanation:

In reactant side: 6 atoms of O, 2 atoms of Cl and 2 atoms of K

In product side: 2 atoms of K and 2 atoms of Cl + 6 atoms of O

CO + 3H₂ → CH₃OH

In reactant side: 1 C, 1 O and 6 H

In product side: 1 C, 4 H and 1 O

PCl₃ + 2Cl₂ → PCl₅

In reactant side: 1 P, 7 Cl

In product side. 1 P, 5 Cl

Mg + N₂ → Mg₃N₂

In reactant side 1 Mg and 2 N

In product side 3 Mg and 2 N

7 0
3 years ago
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