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Fudgin [204]
3 years ago
7

For the reaction C + 2H2 → CH4, how many moles of hydrogen are needed to make 108.6 grams of methane, CH4 ?

Chemistry
1 answer:
dlinn [17]3 years ago
6 0

Answer:

RFM \: of \: methane \:  = 16 \: g \\ 16 \: g  \: are \: weighed \: by \: 1 \: mole \: of \: methane \\ 108.6 \: g \: are \: weighed \: by \: ( \frac{108.6}{16}  \times 1) \: moles \\  = 6.7875 \: moles \: of \: methane \\ from \: the \: equation :  \\ 1 \: mole \: of \: methane \: is \: formed \: by \: 2 \: moles \: of \: hydrogen \\ 6.7875 \: moles \: of \: methane \: will \: be \: produced \: by \: (6.7875 \times 2) \: moles \\  = 13.6 \: moles \: of \: hydrogen

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Snezhnost [94]

answer:

612.5

Explanation:

from the equation, no. of mole of iron oxide is obtaibed from mass over molar mass equals to 5.5 moles, taking mole ratio as 1:2 to get moles of iron as 10.94 moles, then the mass is obtained by moles of iron times molar mass

5 0
2 years ago
How many grams of fluorine must be reacted with excess lithium iodide to produce 10.0 grams of lithium fluoride?
rewona [7]
Answer:
             7.32 g of F₂

Solution:
              The equation is as follow,

                                   2 LiI  +  F₂    →    2 LiF  +  I₂

According to equation,

           51.88 g (2 mole) of LiF is produced from  =  37.99 g (1 mole) F₂
So,
                          10 g of LiF will be produced by  =  X g of F₂

Solving for X,
                      X  =  (10 g × 37.99 g) ÷ 51.88 g

                      X  =  7.32 g of F₂
8 0
3 years ago
Phenolphthalein is used as a visual indicator in a strong acid-strong base titration because
marysya [2.9K]
A strong acid- strong base titration is performed using a phenolphthalein indicator. Phenolphtalein is chosen because it changes color in a pH range between 8.3 – 10. It will appear pink in basic solutions and clear in acidic solutions. ... It is known as the titrant.
3 0
3 years ago
How many joules of heat are required to raise the temperature of 320.0 grams of water from 24.2°C to 50.8°C? Heat capacity of wa
balu736 [363]

Answer:

35750.4 Joules

Explanation:

Using the formula as follows;

Q = m × c × ∆T

Where;

Q = amount of heat (joules)

m = mass of substance (g)

c = specific heat capacity (J/g°C)

∆T = change in temperature (°C)

According to the provided information,

mass (m) = 320.0 grams

c = 4.2 J/g°C

∆T = (50.8°C - 24.2°C) = 26.6°C

Q = ?

Using; Q = m × c × ∆T

Q = 320 × 4.2 × 26.6

Q = 35750.4 J

5 0
3 years ago
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 alcoho
ankoles [38]

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.

Density of the ethanol = 0.785 g/cm³

The volume of laboratory = l × b × h

= 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.

5 0
2 years ago
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