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igor_vitrenko [27]
4 years ago
11

A 0.1149-g sample of solid magnesium is burned in a bomb calorimeter that has a heat capacity of 1769 J/°C. The calorimeter cont

ains exactly 321 g of water, and the temperature increases by 1.126°C. Calculate the heat given off by the burning Mg, in kJ/g and in kJ/mol.
Chemistry
1 answer:
Andreas93 [3]4 years ago
8 0

Answer:

The heat given off by the burning Mg is 30.5 kJ/grams or 745.5 kJ/mol

Explanation:

<u>Step 1: </u>Data given

Mass of magnesium sample = 0.1149 grams

Molar mass of magnesium = 24.3 g/mol

bomb calorimeter that has a heat capacity of 1769 J/°C

Mass of water = 321 grams

Temperature change = 1.126 °C

Step 2: Calculated heat released

Q = (1769 J/°C * 1.126 °C) + (321 grams * 4.184 J/g°C * 1.126 °C)

Q = 3504.18 J =3.504 kJ

<u>Step 3:</u> Calculate the heat given off by the burning Mg, in kJ/g

3.504 kJ / 0.1149 grams = 30.5 kJ/grams

<u>Step 4:</u> Calculate moles of Magnesium

Moles of magnesium = mass magnesium / Molar mass magnesium

Moles magnesium = 0.1149 grams / 24.3 g/mol

Moles magnesium = 0.0047 moles

<u>Step 5:</u> Calculate the heat given off by the burning Mg, in kJ/mol

3.504 kJ / 0.0047 moles = 745.5 kJ/mol

The heat given off by the burning Mg is 30.5 kJ/grams or 745.5 kJ/mol

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An alpha particle is identical to a helium atom that has been stripped of its two electrons
8 0
3 years ago
An alkaline battery produces electrical energy according to the following equation.
Gnesinka [82]

Answer:

Part a: limiting reactant MnO₂

Part b: 12.43 g of Zn(OH)₂

Explanation:

Part a

Data given:

mass of Zn = 37.0 g

mass of MnO₂ = 21.5 g

Limiting reactant = ?

Given Reaction

        Zn(s) + 2 MnO₂(s) + H₂O(l) ---------->Zn(OH)₂(s) + Mn₂O₃(s)

To determine the limiting reactant first we will look at the reaction

         Zn(s)  +  2 MnO₂(s) + H₂O(l) ---------->Zn(OH)₂(s) + Mn₂O₃(s)

          1 mol     2 mol

Convert moles to mass

molar mass of Zn = 65.4 g/mol

Mass of MnO₂ = 55 + 2 (16) = 55 + 32 = 87 g/mol

So,

       Zn(s)          +      2 MnO₂(s)    +    H₂O(l) ----------> Zn(OH)₂(s) + Mn₂O₃(s)

1 mol (65.4 g/mol)    2 mol (87 g/mol)

       65 g                        174 g

So its clear from the reaction that 65 g Zn react with 174 g of MnO₂.

now if we look at the given amounts the amount MnO₂ is less then the amount of Zn but in actual calculation amount of MnO₂ is more then amount of zinc.

So, for MnO₂ if we calculate the needed amount of zinc

So apply unity formula

           65 g Zn react ≅ 174 g of MnO₂

            X g of Zn ≅ 21.5 g of MnO₂

Do cross multiplication

           X g Zn react = 65 g x 21.5 g / 174 g

           X g of Zn ≅ 8.032 g

So, 8.032 g of zinc will react out of 37.0 grams. the remaining will be in excess.

So MnO₂ will be consumed completely an it will be limiting reactant.

____________

part b

Data given:

mass of Zn = 37.0 g

mass of MnO₂ = 21.5 g

Mass of Zn(OH)₂  produced = ?

Given Reaction

        Zn(s) + 2 MnO₂(s) + H₂O(l) ---------->Zn(OH)₂(s) + Mn₂O₃(s)

Solution:

As from the part A we come to know that MnO₂ is limiting reactant, so the amount of Zn(OH)₂ will depend on the amount of MnO₂.

So first we convert mass of MnO₂ to moles

Formula Used

        no. of moles = mass in grams / molar mass

Mass of MnO₂ = 55 + 2 (16)

Mass of MnO₂ = 55 + 32 = 87 g/mol

Put values in the above equation

        no. of moles = 21.5 g / 87 g/mol

        no. of moles = 0.25 moles

Now,

Look at the reaction

         Zn(s)  +  2 MnO₂(s) + H₂O(l) ---------->Zn(OH)₂(s) + Mn₂O₃(s)

                        2 mol                                       1 mol  

So its clear from the reaction that 2 mole of MnO₂ gives 1 mole of Zn(OH)₂

then how many moles of Zn(OH)₂ will be produce by 0.25 moles of MnO₂

So.

apply unity formula

         2 mol of MnO₂ ≅ 1 mole of Zn(OH)₂

            0.25 moles of MnO₂  ≅ X mole of Zn(OH)₂

Do cross multiplication

           X mole of Zn(OH)₂ = 1 mole x 0.25 mol x  / 2 mol

         X mole of Zn(OH)₂  ≅ 0.125

Now Conver moles of  Zn(OH)₂  to mass

Formula used

         mass in grams = no. of moles x molar mass

Molar mass of  Zn(OH)₂

Molar mass of  Zn(OH)₂ = 65.4 + 2 (16 + 1)

Molar mass of  Zn(OH)₂ = 65.4 + 2 (17)

Molar mass of  Zn(OH)₂ = 65.4 + 34 = 99.4 g/mol

Put values in above equation

        mass in grams = 0.125 mol x 99.4 g/mol  

        mass in grams = 12.43 g

So,

12.43 g of Zn(OH)₂  will be produce.

5 0
3 years ago
What volume of nitrogen dioxide is formed at 735 torr and 28.2 °C by reacting 3.56 cm3 of copper (d = 8.95 g/cm3) with 200 mL of
weqwewe [10]

Answer:

25.76 L

Explanation:

Given, Volume of Copper = 3.56 cm³

Density = 8.95 g/cm³

Considering the expression for density as:

Density=\frac {Mass}{Volume}

So,

So, Mass= Density * Volume = 8.95 g/cm³ * 3.56 cm³ = 31.862 g

Mass of copper = 31.862 g

Molar mass of copper = 63.546 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{31.862\ g}{63.546\ g/mol}

<u>Moles of copper = 0.5014 moles </u>

Given, Volume of nitric acid solution = 200 mL = 200 cm³

Density = 1.42 g/cm³

Considering the expression for density as:

Density=\frac {Mass}{Volume}

So,

So, Mass= Density * Volume = 1.42 g/cm³ * 200 cm³ = 284 g

Also, Nitric acid is 68.0 % by mass. So,  

Mass of nitric acid = \frac {68}{100}\times 284\ g = 193.12 g

Molar mass of nitric acid = 63.01 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{193.12\ g}{63.01\ g/mol}

<u>Moles of nitric acid = 3.0649 moles </u>

According to the reaction,  

Cu_{(s)}+4HNO_3_{(aq)}\rightarrow Cu(NO_3)_2_{(aq)} + 2NO_2_{(g)} + 2H_2O_{(l)}

1 mole of copper react with 4 moles of nitric acid

Thus,  

0.5014 moles of copper react with 4*0.5014 moles of nitric acid

Moles of nitric acid required = 2.0056 moles

Available moles of nitric acid = 3.0649 moles

<u>Limiting reagent is the one which is present in small amount. Thus, nitric acid is present in large amount, copper is the limiting reagent. </u>

The formation of the product is governed by the limiting reagent. So,

1 mole of copper on reaction forms 2 moles of nitrogen dioxide

So,

0.5014 mole of copper on reaction forms 2*0.5014 moles of nitrogen dioxide

<u>Moles of nitrogen dioxide = 1.0028 moles </u>

Given:  

Pressure = 735 torr

The conversion of P(torr) to P(atm) is shown below:

P(torr)=\frac {1}{760}\times P(atm)

So,  

Pressure = 735 / 760 atm = 0.9632 atm

Temperature = 28.2 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T₁ = (28.2 + 273.15) K = 301.35 K  

Using ideal gas equation as:

PV=nRT

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

0.9632 atm × V = 1.0028 mol × 0.0821 L.atm/K.mol × 301.35 K  

<u>⇒V = 25.76 L</u>

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