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3 years ago

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What volume of benzene (C6H6, d= 0.88 g/mL, molar mass = 78.11 g/mol) is required to produce 1.5 x 103 kJ of heat according to t

he following reaction?2 C6H6(l) + 15 O2(g) → 12 CO2(g) + 6 H2O(g) ΔH°rxn = -6278 kJ

1 answer:

Anton [14]3 years ago

4 0

Answer:

We need 42.4 mL of benzene to produce 1.5 *10³ kJ of heat

Explanation:

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

Density of benzene = 0.88 g/mL

Molar mass of benzene = 78.11 g/mol

Heat produced = 1.5 * 10³ kJ

<u>Step 2:</u> The balanced equation

2 C6H6(l) + 15 O2(g) → 12 CO2(g) + 6 H2O(g) ΔH°rxn = -6278 kJ

<u>Step 3:</u> Calculate moles of benzen

1.5 * 10³ kJ * (2 mol C6H6 / 6278 kJ) = 0.478 mol C6H6

<u>Step 4:</u> Calculate mass of benzene

Mass benzene : moles benzene * molar mass benzene

Mass benzene= 0.478 * 78.11 g

Mass of benzene = 37.34 grams

<u>Step 5:</u> Calculate volume of benzene

Volume benzene = mass / density

Volume benzene = 37.34 grams / 0.88g/mL

Volume benzene = 42.4 mL

We need 42.4 mL of benzene to produce 1.5 *10³ kJ of heat

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Calculate the mass of methane that must be burned to provide enough heat to convert 242.0 g of water at 26.0°C into steam at 101

Anarel [89]

<u>Answer:</u> The mass of methane burned is 12.4 grams.

<u>Explanation:</u>

The chemical equation for the combustion of methane follows:

$CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H^o_{rxn}=[(1\times \Delta H^o_f_{(CO_2(g))})+(2\times \Delta H^o_f_{(H_2O(g))})]-[(1\times \Delta H^o_f_{(CH_4(g))})+(2\times \Delta H^o_f_{(O_2(g))})]$

We are given:

$\Delta H^o_f_{(H_2O(g))}=-241.82kJ/mol\\\Delta H^o_f_{(CO_2(g))}=-393.51kJ/mol\\\Delta H^o_f_{(CH_4(g))}=-74.81kJ/mol\\\Delta H^o_f_{O_2}=0kJ/mol$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(1\times (-393.51))+(2\times (-241.82))]-[(1\times (-74.81))+(2\times (0))]\\\\\Delta H^o_{rxn}=-802.34kJ$

The heat calculated above is the heat released for 1 mole of methane.

The process involved in this problem are:

$(1):H_2O(l)(26^oC)\rightarrow H_2O(l)(100^oC)\\\\(2):H_2O(l)(100^oC)\rightarrow H_2O(g)(100^oC)\\\\(3):H_2O(g)(100^oC)\rightarrow H_2O(g)(101^oC)$

Now, we calculate the amount of heat released or absorbed in all the processes.

<u>For process 1:</u>

$q_1=mC_p,l\times (T_2-T_1)$

where,

$q_1$ = amount of heat absorbed = ?

m = mass of water = 242.0 g

$C_{p,l}$ = specific heat of water = 4.18 J/g°C

$T_2$ = final temperature = $100^oC$

$T_1$ = initial temperature = $26^oC$

Putting all the values in above equation, we get:

$q_1=242.0g\times 4.18J/g^oC\times (100-(26))^oC=74855.44J$

<u>For process 2:</u>

$q_2=m\times L_v$

where,

$q_2$ = amount of heat absorbed = ?

m = mass of water or steam = 242 g

$L_v$ = latent heat of vaporization = 2257 J/g

Putting all the values in above equation, we get:

$q_2=242g\times 2257J/g=546194J$

<u>For process 3:</u>

$q_3=mC_p,g\times (T_2-T_1)$

where,

$q_3$ = amount of heat absorbed = ?

m = mass of steam = 242.0 g

$C_{p,g}$ = specific heat of steam = 2.08 J/g°C

$T_2$ = final temperature = $101^oC$

$T_1$ = initial temperature = $100^oC$

Putting all the values in above equation, we get:

$q_3=242.0g\times 2.08J/g^oC\times (101-(100))^oC=503.36J$

Total heat required = $q_1+q_2+q_3=(74855.44+546194+503.36)=621552.8J=621.552kJ$

To calculate the number of moles of methane, we apply unitary method:

When 802.34 kJ of heat is needed, the amount of methane combusted is 1 mole

So, when 621.552 kJ of heat is needed, the amount of methane combusted will be = $\frac{1}{802.34}\times 621.552=0.775mol$

To calculate the number of moles, we use the equation:

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

Molar mass of methane = 16 g/mol

Moles of methane = 0.775 moles

Putting values in above equation, we get:

$0.775mol=\frac{\text{Mass of methane}}{16g/mol}\\\\\text{Mass of methane}=(0.775mol\times 16g/mol)=12.4g$

Hence, the mass of methane burned is 12.4 grams.

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3 years ago

Name two "Storage Polysaccharides" and two "Structural Polysaccharides"

Vsevolod [243]

Answer:

Examples of storage polysaccharides - <u>starch and glycogen</u> and structural polysaccharides - <u>cellulose and chitin</u>

Explanation:

Polysaccharides are the complex carbohydrate polymers, composed of monosaccharide units that are joined together by glycosidic bond.

In other words, polysaccharides are the carbohydrate molecules that give monosaccharides or oligosaccharides on hydrolysis.

The examples of storage polysaccharides are starch and glycogen. The examples of structural polysaccharides are cellulose and chitin.

6 0

3 years ago

one method for generating chlorine gas is by reacting potassium permanganate and hydrochloric acid. how many liters of Cl2 at 40

Ronch [10]

<u>Answer:</u> The volume of chlorine gas produced in the reaction is 2.06 L.

<u>Explanation:</u>

<u>For potassium permanganate:</u>

To calculate the number of moles, we use the equation:

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

Given mass of potassium permanganate = 6.23 g

Molar mass of potassium permanganate = 158.034 g/mol

Putting values in above equation, we get:

$\text{Moles of potassium permanganate}=\frac{6.23g}{158.034g/mol}=0.039mol$

<u>For hydrochloric acid:</u>

To calculate the moles of hydrochloric acid, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Molarity of HCl = 6.00 M

Volume of HCl = 45.0 mL = 0.045 L (Conversion factor: 1 L = 1000 mL)

Putting values in above equation, we get:

$6.00mol/L=\frac{\text{Moles of HCl}}{0.045L}\\\\\text{Moles of HCl}=0.27mol$

For the reaction of potassium permanganate and hydrochloric acid, the equation follows:

$2KMnO_4+16HCl\rightarrow 2MnCl_2+5Cl_2+2KCl+8H_2O$

By Stoichiometry of the reaction:

16 moles of hydrochloric acid reacts with 2 moles of potassium permanganate.

So, 0.27 moles of hydrochloric acid will react with = $\frac{2}{16}\times 0.27=0.033moles$ of potassium permanganate.

As, given amount of potassium permanganate is more than the required amount. So, it is considered as an excess reagent.

Thus, hydrochloric acid is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

16 moles of hydrochloric acid reacts with 5 moles of chlorine gas.

So, 0.27 moles of hydrochloric acid will react with = $\frac{5}{16}\times 0.27=0.0843moles$ of chlorine gas.

To calculate the volume of gas, we use the equation given by ideal gas equation:

$PV=nRT$

where,

P = pressure of the gas = 1.05 atm

V = Volume of gas = ? L

n = Number of moles = 0.0843 mol

R = Gas constant = $0.0820\text{ L atm }mol^{-1}K^{-1}$

T = temperature of the gas = $40^oC=[40+273]K=313K$

Putting values in above equation, we get:

$1.05atm\times V=0.0843\times 0.0820\text{ L atm }mol^{-1}K^{-1}\times 313K\\\\V=2.06L$

Hence, the volume of chlorine gas produced in the reaction is 2.06 L.

3 0

4 years ago

Which statements about freshwater sources are true? Check all that apply.

sammy [17]

Answer:

Explanation:

1 is true

2 is false

3 is true

4 is true

5 is false

That is what I got from researching.

4 0

3 years ago

Read 2 more answers

Given the following information, what is the concentration of H2O(g) at equilibrium? [H2S](eq) = 0.671 M [O2](eq) = 0.587 M Kc =

MAVERICK [17]

<u>Answer:</u> The equilibrium concentration of water is 0.597 M

<u>Explanation:</u>

Equilibrium constant in terms of concentration is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as $K_{c}$

For a general chemical reaction:

$aA+bB\rightleftharpoons cC+dD$

The expression for $K_{eq}$ is written as:

$K_{c}=\frac{[C]^c[D]^d}{[A]^a[B]^b}$

The concentration of pure solids and pure liquids are taken as 1 in the expression.

For the given chemical reaction:

$2H_2S(g)+O_2(g)\rightleftharpoons 2S(s)+2H_2O(g)$

The expression of $K_c$ for above equation is:

$K_c=\frac{[H_2O]^2}{[H_2S]^2\times [O_2]}$

We are given:

$[H_2S]_{eq}=0.671M$

$[O_2]_{eq}=0.587M$

$K_c=1.35$

Putting values in above expression, we get:

$1.35=\frac{[H_2O]^2}{(0.671)^2\times 0.587}$

$[H_2O]=\sqrt{(1.35\times 0.671\times 0.671\times 0.587)}=0.597M$

Hence, the equilibrium concentration of water is 0.597 M

8 0

4 years ago