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3 years ago

15

Liquid octane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 17. g of octane is mi

xed with 93.0 g of oxygen. Calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction. Be sure your answer has the correct number of significant digits.

1 answer:

Lisa [10]3 years ago

8 0

Answer: 52.8 g of $CO_2$

Explanation:

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}$

a) moles of octane

$\text{Number of moles}=\frac{17.0g}{114g/mol}=0.150moles$

b) moles of oxygen

$\text{Number of moles}=\frac{93.0g}{32g/mol}=2.91moles$

$2C_8H_{18}+25O_2\rightarrow 16CO_2+18H_2O$

According to stoichiometry :

2 moles of $C_8H_{18}$ require 25 moles of $O_2$

Thus 0.150 moles of $C_8H_{18}$ require= $\frac{25}{2}\times 0.150=1.875moles$ of $O_2$

Thus $C_8H_{18}$ is the limiting reagent as it limits the formation of product and $O_2$ is the excess reagent.

As 2 moles of $C_8H_{18}$ give = 16 moles of $CO_2$

Thus 0.150 moles of $C_8H_{18}$ give = $\frac{16}{2}\times 0.150=1.2moles$ of $CO_2$

Mass of $CO_2=moles\times {\text {Molar mass}}=1.2moles\times 44g/mol=52.8g$

Thus 52.8 g of $CO_2$ will be produced from the given masses of both reactants.

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When 1.045 g of CaO is added to 50.0 mL of water at 25.0 °C in a calorimeter, the temperature of the water increases to 32.3 °C.

Rus_ich [418]

Answer:

1.71 kJ/mol

Explanation:

The expression for the calculation of the enthalpy change of a process is shown below as:-

$\Delta H=m\times C\times \Delta T$

Where,

$\Delta H$ is the enthalpy change

m is the mass

C is the specific heat capacity

$\Delta T$ is the temperature change

Thus, given that:-

Mass of CaO = 1.045 g

Specific heat = 4.18 J/g°C

$\Delta T=32.3-25.0\ ^0C=7.3\ ^0C$

So,

$\Delta H=1.045\times 4.18\times 7.3\ J=31.88713\ J$

Also, 1 J = 0.001 kJ

So,

$\Delta H=0.03189\ kJ$

Also, Molar mass of CaO = 56.0774 g/mol

$Moles=\frac{Mass}{Molar\ mass}=\frac{1.045}{56.0774}\ mol=0.01863\ mol$

Thus, Enthalpy change in kJ/mol is:-

$\Delta H=\frac{0.03189}{0.01863}\ kJ/mol=1.71\ kJ/mol$

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3 years ago

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Vanyuwa [196]

Answer:

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Explanation:

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3 years ago

A piece of wood has a labeled length value of 63.2 cm. You measure its length three times and record the following data: 63.1 cm

Ulleksa [173]

Percent error (%)= $\frac{\left | Accepted value - Measured value \right |}{Accepted value}\times 100$

Accepted value is true value.

Measured values is calculated value.

In the question given Accepted value (true value) = 63.2 cm

Given Measured(calculated values) = 63.1 cm , 63.0 cm , 63.7 cm

1) Percent error (%) for first measurement.

Accepted value (true value) = 63.2 cm, Measured(calculated values) = 63.1 cm

Percent error (%)= $\frac{\left | Accepted value - Measured value \right |}{Accepted value}\times 100$

$Percent error = \frac{\left | 63.2 - 63.1 \right |}{63.2}\times 100$

$Percent error = \frac{0.1}{63.2}\times 100$

$Percent error = 0.00158\times 100$

Percent error = 0.158 %

2) Percent error (%) for second measurement.

Accepted value (true value) = 63.2 cm, Measured(calculated values) = 63.0 cm

Percent error (%)= $\frac{\left | Accepted value - Measured value \right |}{Accepted value}\times 100$

$Percent error = \frac{\left | 63.2 - 63.0 \right |}{63.2}\times 100$

$Percent error = \frac{0.2}{63.2}\times 100$

$Percent error = 0.00316\times 100$

Percent error = 0.316 %

3) Percent error (%) for third measurement.

Accepted value (true value) = 63.2 cm, Measured(calculated values) = 63.7 cm

Percent error (%)= $\frac{\left | Accepted value - Measured value \right |}{Accepted value}\times 100$

$Percent error = \frac{\left | 63.2 - 63.7 \right |}{63.2}\times 100$

$Percent error = \frac{\left | -0.5 \right |}{63.2}\times 100$

$Percent error = \frac{(0.5)}{63.2}\times 100$

$Percent error = 0.00791\times 100$

Percent error = 0.791 %

Percent error for each measurement is :

63.1 cm = 0.158%

63.0 cm = 0.316%

63.7 cm = 0.791%

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4 years ago

If 600.0 ml of air at 20.0 c is heated to 60.0L , what will be its volume

Deffense [45]

P1v1/t1 = p2t2/t2

p is constant
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