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Y_Kistochka [10]
3 years ago
6

If the density of ethanol, C2H5OH, is 0.789 g/mL. How many milliliters of ethanol are needed to produce 15.0 g of CO2 according

to the following chemical equation? C2H5OH(l) + 3 O2(g) → 2 CO2(g) + 3 H2O(l)
Chemistry
2 answers:
Svetradugi [14.3K]3 years ago
7 0

Answer:

9.94 mL, the volume of ethanol needed

Explanation:

The reaction is:

C₂H₅OH(l) + 3 O₂(g) → 2 CO₂(g) + 3 H₂O(l)

We convert the mass of the formed product to moles:

15 g . 1mol / 44g = 0.341 moles

2 moles of dioxide are produced by 1 mol of ethanol, in order to stoichiometry.

Therefore, 0.341 moles of CO₂ must be produced by (0.341. 1) / 2 = 0.1705 moles of alcohol.

We convert the moles to mass, and then, the mass to volume by the use of density.

0.1705 mol . 46 g / 1 mol = 7.84 g of ethanol

Ethanol density = Ethanol mass /Ethanol volume

Ethanol volume = Ethanol mass /Ethanol density → 7.84 g /0.789 g/mL =

9.94 mL

ad-work [718]3 years ago
6 0

Answer:

We need 9.95 mL of ethanol

Explanation:

Step 1: Data given

Density ethanol = 0.789 g/mL

Mass CO2 produced = 15.0 grams

Step 2: The balanced equation

C2H5OH(l) + 3 O2(g) → 2 CO2(g) + 3 H2O(l)

Step 3: Calculate moles CO2

Moles CO2 = mass CO2 / molar mass CO2

Moles CO2 = 15.0 grams / 44.01 g/mol

Moles CO2 = 0.341 moles

Step 4: Calculate moles ethanol

For 1 mol ethanol, 3 moles we need O2 to produce 2 moles CO2 and 3 moles H2O

For 0.341 moles CO2 we need 0.341 /2 = 0.1705 moles ethanol

Step 5: Calculate mass ethanol

Mass ethanol = moles ethanol * molar mass ethanol

Mass ethanol = 0.1705 moles * 46.07 g/mol

Mass ethanol = 7.85 grams

Step 6: Calculate volume ethanol

Volume ethanol = mass / density

Volume = 7.85 grams / 0.789 g/mL

Volume= 9.95 mL

We need 9.95 mL of ethanol

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Masja [62]

Answer:

When a gas occupies a smaller volume, it exerts a higher pressure; when it occupies a larger volume, it exerts a lower pressure (assuming the amount of gas and the temperature do not change). Since P and V are inversely proportional, a graph of 1/P vs. V is linear.

Explanation:

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6 0
3 years ago
A 1.000 g sample of an unknown hydrate of cobalt chloride is gently dehydrated. The resulting mass is 0.546 g. The cobalt is iso
ira [324]

Answer:

Explanation:

From the 1 g sample you have:

0.546 grams of cobalt chloride

1-0.546=0.454 grams of water

Now:

1) The salt

Of the 0.546 g, 0.248 g are cobalt (Mr=58.9) and the rest id Cl (Mr=35.45):

n_{Co}=\frac{0.248 g}{58.9 g/mol}=4.21*10^{-3}mol

n_{Cl}=\frac{0.298 g}{35.45g/mol}=8.4*10^{-3}mol

Dividing:

\frac{n_{Cl}}{n_{Co}}=\frac{8.4*10^{-3}mol}{4.21*10^{-3}mol}=2

So the molecular formula will be:

CoCl_2

2) The water

The water's molecular weight is M=18 :

n_{w}=\frac{0.454g}{18g/mol}=0.025 mol

Bonding with the Co:

\frac{n_{w}}{n_{Co}}=\frac{0.025mol}{4.21*10^{-3}mol}=6

The complete formula of the hydrate:

CoCl_2*6 H_2O

8 0
4 years ago
Calculate Delta G for each reaction using Delta Gf values: answer kJ ...thank you
Leni [432]

Answer:

a) \Delta G=2.6kJ

b) \Delta G=-979.57kJ

c) \Delta G=264.21kJ

Explanation:

Hello,

In this case, in each reaction we must subtract the Gibbs free energy of formation the reactants to the Gibbs free energy of formation of the products considering each species stoichiometric coefficients. In such a way, the Gibbs free energy of formations are:

\Delta _fG_{H_2}=\Delta _fG_{I_2}=0kJ/mol\\\Delta _fG_{HI}=1.3kJ/mol\\\Delta _fG_{CO_2}=-394.4kJ/mol\\\Delta _fG_{CO}=-137.3 kJ/mol\\\Delta _fG_{NH_3}=16.7 kJ/mol\\\Delta _fG_{HCl}=-95.3kJ/mol\\\Delta _fG_{MnO_2}=465.37kJ/mol\\\Delta _fG_{Mn}=0kJ/mol\\\Delta _fG_{NH_4Cl}=-342.81kJ/mol

So we proceed as follows:

a)

\Delta G=2\Delta _fG_{HI}-\Delta _fG_{H_2}-\Delta _fG_{I_2}\\\\\Delta G=2*1.3\\\\\Delta G=2.6kJ

b)

\Delta G=\Delta _fG_{Mn}+2*\Delta _fG_{CO_2}-\Delta _fG_{MnO_2}-2*\Delta _fG_{CO}\\\\\Delta G=0+2*-394.4-465.37-2*-137.3\\\\\Delta G=-979.57kJ

c)

\Delta G=\Delta _fG_{NH_3}+\Delta _fG_{HCl}-\Delta _fG_{NH_4Cl}\\\\\Delta G=16.7-95.3-(-342.81)\\\\\Delta G=264.21kJ

Regards.

6 0
3 years ago
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Alekssandra [29.7K]
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