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Natali5045456 [20]
3 years ago
10

25.5 g of methane burns in the combustion reaction below. How much water is produced in this reaction

Chemistry
1 answer:
gulaghasi [49]3 years ago
5 0
CH4 + O2 ===> H2O + CO2 Bare Equation. This is a Combustion Reaction.
CH4 + 2O2====>2H2O + CO2 Balanced equation. You start with this.

Step One
=======
Find the moles of Methane.
1 mole of methane has a mass of
C = 12
O2 = 2 * 16 = 32
1 mol of  CO2 = 12 + 32 = 44 grams / mole

Step Two 
======
Calculate the number of moles of methane.
given mass (g) = 25.5 grams
molar Mass = 44

mol = given mass / molar mass
mol = 25.5 / 44
mol = 0.5795 mols methane.

Step Three
========
Find the moles of water.
For ever mole of methane, 2 moles of water are produced.

1 mol methane. . . . . . .. .0.5795 mols methane
=========== . . . . .= .  .================ . . . . . 
2 mol water  . . . . . . . . . . .. .. . .x

Mole of water = 2 * 0.5795
mole of water = 1.159 

Step Four
=======
Find the mass of water.
1 mol = 2H + O = 2 + 16 = 18 grams / mol

desired mass = mols * Molar mass
desired mass = 1.159 * 18
desired mass = 20.86 grams of water.

If you do these questions by dimensional analysis, this problem can be done all in 1 step. It would look like this.

25.5 g [1mol /44grams]*[2mol water / 1 mol meth]*[18 grams / 1 mol water]


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If you place 1.0 L of ethanol (C2H5OH) in a small laboratory that is 3.0 m long, 2.0 m wide, and 2.0 m high, will all the alcohol evaporate? If some liquid remains, how much will there be? The vapor pressure of ethyl alcohol at 25 °C is 59 mm Hg, and the density of the liquid at this temperature is 0.785g/cm^3 .

will all the alcohol evaporate? or none at all?

Answer:

Yes, all the ethanol present in the laboratory will evaporate since the mole of ethanol present in vapor is greater. The volume of ethanol left will therefore  be zero.

Explanation:

Given that:

The volume of alcohol which is placed in a small laboratory = 1.0 L

Vapor pressure of ethyl alcohol  at 25 ° C = 59 mmHg

Converting 59 mmHg to atm ; since 1 atm = 760 mmHg;

Then, we have:

= \frac{59}{760}atm

= 0.078 atm

Temperature = 25 ° C

= ( 25 + 273 K)

= 298 K.

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= 3.0 m × 2.0 m × 2.5 m

= 15 m³

Converting the volume of laboratory to liter;

since 1 m³ = 100 L; Then, we  have:

15 × 1000 = 15,000 L

Using ideal gas equation to determine the moles of ethanol in vapor phase; we have:

PV = nRT

Making n the subject of the formula; we have:

n = \frac{PV}{RT}

n = \frac{0.078 * 15000}{0.082*290}

n = 47. 88 mol of ethanol

Moles of ethanol in 1.0 L bottle can be calculated as follows:

Since  numbers of moles = \frac{mass}{molar mass}

and mass = density × vollume

Then; we can say ;

number of moles = \frac{density*volume }{molar mass of ethanol}

number of moles =\frac{0.785g/cm^3*1000cm^3}{46.07g/mol}

number of moles = \frac{&85}{46.07}

number of moles = 17.039 mol

Thus , all the ethanol present in the laboratory will evaporate since the mole of ethanol present in vapor is greater. The volume of ethanol left will therefore be zero.

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The low-resolution mass spectrum of an unknown analgesic X had a molecular ion of 151. Possible molecular formulas include C7H5N
garik1379 [7]

Answer:

The molecular formula of X is C8H9NO2

Explanation:

Step 1: Data given

exact mass of 151.0640

Molar mass of C = 12 g/mol

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Molar mass of O = 15.9949 g/mol

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Step 2: Calculate molar mass of C7H5NO3

7*12 + 5*1.00783 + 14.0031 + 3*15.9949 = 151.02695 g/mol

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