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victus00 [196]
3 years ago
12

The camel stores the fat tristearin (C57H110O6) in its hump. As well as being a source of energy, the fat is also a source of wa

ter because, when it is used, the reaction 2 C57H110O6(g) + 163 O2(g) 114 CO2(g) + 110 H2O(ℓ) takes place. What mass of water is available from 2.2 pound of this fat? Answer in units of g.
Chemistry
1 answer:
Aneli [31]3 years ago
8 0

Answer:

1109 g H₂O

Explanation:

2.2 pounds can be converted to grams using a conversion ratio:

(2.2lb)(453.592g/lb) = 997.9024 g C₅₇H₁₁₀O₆

The mass in grams is converted to moles using the molecular weight of tristearin (891.48 g/mol)

(997.9024 g)(mol/891.48g) = 1.119...mol C₅₇H₁₁₀O₆

The moles of C₅₇H₁₁₀O₆ can be related to the moles of water through the molar ratio:

(1.119mol C₅₇H₁₁₀O₆)(110 H₂O/2 C₅₇H₁₁₀O₆) = 61.545 mol H₂O

The mass of water is then calculated using the molecular weight (18.02 g/mol):

(61.545 mol)(18.02 g/mol) = 1109 g H₂O

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Consider the following reaction: CO(g)+2H2(g)⇌CH3OH(g) Kp=2.26×104 at 25 ∘C. Calculate ΔGrxn for the reaction at 25 ∘C under eac
valentinak56 [21]

Answer : The value of \Delta G_{rxn} is, 8.867kJ/mole

Explanation :

The formula used for \Delta G_{rxn} is:

\Delta G_{rxn}=\Delta G^o+RT\ln Q   ............(1)

where,

\Delta G_{rxn} = Gibbs free energy for the reaction

\Delta G_^o =  standard Gibbs free energy

R = gas constant = 8.314 J/mole.K

T = temperature = 25^oC=273+25=298K

Q = reaction quotient

First we have to calculate the \Delta G_^o.

Formula used :

\Delta G^o=-RT\times \ln K_p

Now put all the given values in this formula, we get:

\Delta G^o=-(8.314J/mole.K)\times (298K)\times \ln (2.26\times 10^{4})

\Delta G^o=-24839.406J/mole=-24.83\times 10^3J/mole=-24.83kJ/mole

Now we have to calculate the value of 'Q'.

The given balanced chemical reaction is,

CO(g)+2H_2(g)\rightarrow CH_3OH(g)

The expression for reaction quotient will be :

Q=\frac{(p_{CH_3OH})}{(p_{CO})\times (p_{H_2})^2}

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.

Now put all the given values in this expression, we get

Q=\frac{(1.4)}{(1.2\times 10^{-2})\times (1.2\times 10^{-2})^2}

Q=8.1\times 10^{5}

Now we have to calculate the value of \Delta G_{rxn} by using relation (1).

\Delta G_{rxn}=\Delta G^o+RT\ln Q

Now put all the given values in this formula, we get:

\Delta G_{rxn}=-24.83kJ/mole+(8.314\times 10^{-3}kJ/mole.K)\times (298K)\ln (8.1\times 10^{5})

\Delta G_{rxn}=8.867\times 10^3J/mole=8.867kJ/mole

Therefore, the value of \Delta G_{rxn} is, 8.867kJ/mole

3 0
3 years ago
An ideal gas in a 1.25-gallon container is at a temperature of 125 degrees Celsius and pressure of 2.5 atmospheres. If the gas i
Tresset [83]
You will need the equation PV = nRT

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V = Volume in L
n = moles
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T = Temperature in Kelvin 

First convert 2.5 atm into kPa:

2.5 X 101.3 = 253.25 kPa

Convert 125 Celsius into Kelvin:

125 + 273 = 398 K

Convert Gallons to Litres:

1.25  X 3.79 = 4.74 L

Plug your values into the equation to solve for n:

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n = (253.25)(4.74)/(8.314)(398)

n = 0.362 moles

Now use M = m/n to solve for the mass of O2

M = Molar Mass 
M = mass
n= moles

32 = m/(0.362)

m = (32)(0.362) 

m = 11.58g
8 0
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