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3 years ago

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A student in chemistry 150-02 weighed out 55.5 g of octane (C8H18) and allowed it to react with oxygen, O2. The products formed

were carbon dioxide (CO2) and water (H2O).
Write a balanced equation for the reaction

How many grams of oxygen are required to react with 55.0g of octane (C8H18)?

How many grams of CO2 are produced from 55.0g of octane (C8H18)?

How many molecules of H2O are produced from 55.0g of octane (C8H18)?

How many gams of C8H18 are required to produce 30.0g of water (H2O).?

1 answer:

Anni [7]3 years ago

5 0

Answer:

Explanation:

Given data:

Mass of octane = 55.5 g

Balanced chemical equation = ?

Mass of oxygen required to react = ?

Mass of CO₂ for med = ?

Molecules of water produced = ?

Mass of octane required to produced 30.0 g of water = ?

Solution:

1)

Chemical equation:

2C₈H₁₈ + 25O₂ → 16CO₂ + 18H₂O

2)

Mass of oxygen required to react = ?

Mass of octane = 55.0 g

Solution:

Number of moles of octane:

Number of moles = mass/ molar mass

Number of moles = 55.0 g/114.23 g/mol

Number of moles = 0.48 mol

Now we will compare the moles of octane with oxygen.

C₈H₁₈ : O₂

2 : 25

0.48 : 25/2×0.48 = 6 mol

Mass of oxygen required:

Mass = number of moles × molar mass

Mass = 6 mol × 32 g/mol

Mass = 192 g

3)

Given data:

Mass of carbon dioxide produced = ?

Mass of octane = 55g

Solution:

Number of moles of octane:

Number of moles = mass/ molar mass

Number of moles = 55.0 g/114.23 g/mol

Number of moles = 0.48 mol

Now we will compare the moles of octane with CO₂.

C₈H₁₈ : CO₂

2 : 16

0.48 : 16/2×0.48 = 3.84 mol

Mass of CO₂ produced:

Mass = number of moles × molar mass

Mass = 3.84 mol × 44 g/mol

Mass = 168.96 g

4)

Given data:

Molecules of water produced = ?

Mass of octane = 55g

Solution:

Number of moles of octane:

Number of moles = mass/ molar mass

Number of moles = 55.0 g/114.23 g/mol

Number of moles = 0.48 mol

Now we will compare the moles of octane with H₂O.

C₈H₁₈ : H₂O

2 : 18

0.48 : 18/2×0.48 = 4.32 mol

Number of molecules of water:

1 mol = 6.022× 10²³ molecules

4.32 mol × 6.022× 10²³ molecules/ 1 mol

26 × 10²³ molecules

5)

Given data:

Mass of octane required = ?

Mass of water produced = 30 g

Solution:

Number of moles of water.

Number of moles = mass/ molar mass

Number of moles = 30 g/ 18 gmol

Number of moles = 1.67 mol

Now we will compare the moles of water and octane from balance chemical equation:

2C₈H₁₈ + 25O₂ → 16CO₂ + 18H₂O

H₂O : C₈H₁₈

18 : 2

1.67 : 2/18×1.67 = 0.185 mol

Mass of octane:

Mass = number of moles ×molar mass

Mass = 0.185 × 114.23 g/mol

Mass = 21.13 g

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vagabundo [1.1K]

Answer:

0.712 moles of NO₂ are formed.

Explanation:

First, we need to write the balanced equation:

2 N₂O₅(g) ⇄ 4 NO₂(g) + O₂(g)

From the balanced equation, we can see the relationship between the moles of N₂O₅ and the moles of NO₂. Every 2 moles of N₂O₅ that react, 4 moles of NO₂ are formed. Let us apply this relationship to the information given by the problem (0.356 moles of N₂O₅):

$0.356molN_{2}O_{5}.\frac{4molNO_{2}}{2molN_{2}O_{5}} =0.712molNO_{2}$

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3 years ago

Nicotine, a component of tobacco, is composed of C, H, and N. A 7.875-mg sample of nicotine was combusted, producing 21.363 mg o

Gnom [1K]

Answer: The empirical formula for the given compound is $C_5H_7N$

Explanation:

The chemical equation for the combustion of compound having carbon, hydrogen, and nitrogen follows:

$C_xH_yN_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of carbon, hydrogen and nitrogen respectively.

We are given:

Mass of $CO_2=21.363mg=21.363\times 10^3g=21363g$

Mass of $H_2O=6.125g=6.125\times 10^3g=6125g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 21363 g of carbon dioxide, $\frac{12}{44}\times 21363=5826.27g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in 6125 g of water, $\frac{2}{18}\times 6125=680.55$ of hydrogen will be contained.

Now we have to calculate the mass of nitrogen.

Mass of nitrogen in the compound = (7875) - (5826.27 + 680.55) = 1368.18 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{5826.27g}{12g/mole}=485.52moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{680.55g}{1g/mole}=680.55moles$

Moles of Nitrogen = $\frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{1368.18g}{14g/mole}=97.73moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.0154 moles.

For Carbon = $\frac{485.52}{97.73}=4.96\approx 5$

For Hydrogen = $\frac{680.55}{97.73}=6.96\approx 7$

For Nitrogen = $\frac{97.73}{97.73}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : N = 5 : 7 : 1

Hence, the empirical formula for the given compound nicotine is $C_5H_7N_1=C_5H_7N$

7 0

3 years ago

What is the engine piston displacement in liters of an engine whose displacement is listed as 490 in^3?

marishachu [46]

Answer:

490 in^3 = 8.03 L

Explanation:

Given:

The engine displacement = 490 in^3

= 490 in³

To determine the engine piston displacement in liters L;

(NOTE: Both in^3 (in³) and L are units of volume). Hence, to find the engine piston displacement in liters (L), we will convert in^3 to liters (L)

First, we will convert in³ to cm³

Since 1 in = 2.54 cm

∴ 1 in³ = 16.387 cm³

If 1 in³ = 16.387 cm³

Then 490 in³ = (490 in³ × 16.387 cm³) / 1 in³ = 8029.63 cm³

∴ 490 in³ = 8029.63 cm³

Now will convert cm³ to dm³

(NOTE: 1 L = 1 dm³)

1 cm = 1 × 10⁻² m = 1 × 10⁻¹ dm

∴ 1 cm³ = 1 × 10⁻⁶ m³ = 1 × 10⁻³ dm³

If 1 cm³ = 1 × 10⁻³ dm³

Then, 8029.63 cm³ = (8029.63 cm³ × 1 × 10⁻³ dm³) / 1 cm³ = 8.02963 dm³

≅ 8.03 dm³

∴ 8029.63 cm³ = 8.03 dm³

Hence, 490 in³ = 8029.63 cm³ = 8.03 dm³

Since 1L = 1 dm³

∴ 8.03 dm³ = 8.03 L

Hence, 490 in³ = 8.03 L

3 0

3 years ago

The percent by mass of bicarbonate (HCO3−) in a certain Alka-Seltzer product is 32.5 percent. Calculate the volume of CO2 genera

nevsk [136]

Answer:

The volume of carbon dioxide gas generated 468 mL.

Explanation:

The percent by mass of bicarbonate in a certain Alka-Seltzer = 32.5%

Mass of tablet = 3.45 g

Mass of bicarbonate = $3.45 g\times \frac{32.5}{100}=1.121 mol$

Moles of bicarbonate ion = $\frac{1.121 g/mol}{61 g/mol}=0.01840 mol$

$HCO_3^{-}(aq)+HCl(aq)\rightarrow H_2O(l)+CO_2(g)+Cl^-(aq)$

According to reaction, 1 mole of bicarbonate ion gives with 1 mole of carbon dioxide gas , then 0.01840 mole of bicarbonate ion will give:

$\frac{1}{1}\times 0.01840 mol=0.01840 mol$ of carbon dioxide gas

Moles of carbon dioxide gas n = 0.01840 mol

Pressure of the carbon dioxide gas = P = 1.00 atm

Temperature of the carbon dioxide gas = T = 37°C = 37+273 K=310 K

Volume of the carbon dioxide gas = V

$PV=nRT$ (ideal gas equation)

$V=\frac{nRT}{P}=\frac{0.01840 mol\times 0.0821 atm L/mol K\times 310 K}{1.00 atm}=0.468 L$

1 L = 1000 mL

0.468 L =0.468 × 1000 mL = 468 mL

The volume of carbon dioxide gas generated 468 mL.

5 0

3 years ago

Many computer chips are manufactured from silicon, which occurs in nature as SiO2. When SiO2 is heated to melting, it reacts wit

riadik2000 [5.3K]

Answer:

A) SiO2 is the limiting reactant

B) Theoretical yield= 72333.3g

C) % yield =91.5%

Explanation:

SiO2(s) + 2C(s) --------------> Si(s) + 2CO(g)

n(SiO2)= 155000/60 = 2583.33 mols

n(C)= 79000/12= 3291.66 mols

a)SiO2 is the limiting reactant

According to the balanced reaction equation,

60g of SiO2 produced 28g of SiO2

155000g of SiO2 will produce 155000×28/60= 72333.3g

Therefore theoretical yield of Si= 72333.3g

% yield= 66200/72333.3×100/1 =91.5%

5 0

3 years ago