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Hitman42 [59]
3 years ago
5

From the following balanced equation, CH4(g)+2O2(g)⟶CO2(g)+2H2O(g) how many grams of H2O can be formed when 1.25g CH4 are combin

ed with 1.25×10^23 molecules O2? Use 6.022×10^23 mol−1 for Avogadro's number.
Chemistry
1 answer:
Montano1993 [528]3 years ago
7 0

Answer:

2.81 g of H2O.

Explanation:

We'll begin by calculating mass of O2 that contains 1.25×10²³ molecules O2.

This can be obtained as follow:

From Avogadro's hypothesis, we understood that 1 mole of any substance contains 6.022×10²³ molecules. This implies that 1 mole of O2 also contains 6.022×10²³ molecules.

1 mole of O2 = 16x2 = 32 g.

Thus 6.022×10²³ molecules is present in 32 g of O2,

Therefore, 1.25×10²³ molecules will be present in =

(1.25×10²³ × 32) / 6.022×10²³ = 6.64 g of O2.

Therefore, 1.25×10²³ molecules present in 6.64 g of O2.

Next, the balanced equation for the reaction. This is given below:

CH4(g) + 2O2(g) —> CO2(g) + 2H2O(g)

Next, we shall determine the masses of CH4 and O2 that reacted and the mass of H2O produced from the balanced equation.

This can be obtained as follow:

Molar mass of CH4 = 12 + (4x1) = 16 g/mol.

Mass of CH4 from the balanced equation = 1 x 16 = 16 g

Molar mass of O2 = 16x2 = 32 g/mol.

Mass of O2 from the balanced equation = 2 x 32 = 64 g

Molar mass of H2O = (2x1) + 16 = 18 g/mol.

Mass of H2O from the balanced equation = 2 x 18 = 36 g

From the balanced equation above,

16 g of CH4 reacted with 64 g of O2 to produce 36 g if H2O.

Next, we shall determine the limiting reactant.

This can be obtained as follow:

From the balanced equation above,

16 g of CH4 reacted with 64 g of O2.

Therefore, 1.25 g of CH4 will react with = (1.25 x 64)/16 = 5 g of O2.

From the above calculations, we can see that only 5 g out of 6.64 g of O2 is needed to react completely with 1.25 g of CH4.

Therefore, CH4 is the limiting reactant.

Finally, we shall determine the mass of H2O produced from the reaction.

In this case, the limiting reactant will be used because it will give the maximum yield of H2O.

The limiting reactant is CH4 and the mass of H2O produced from the reaction can be obtained as follow:

From the balanced equation above,

16 g of CH4 reacted to produce produce 36 g if H2O.

Therefore, 1.25 g of CH4 will react to produce = (1.25 x 36)/16 = 2.81 g of H2O.

Therefore, 2.81 g of H2O were obtained from the reaction.

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Select the coefficients necessary to balance each equation. Choose a coefficient for every compound.
ivolga24 [154]

Answer:

See below

Explanation:

2C2H6 + 7O2 = 4CO2 + 6H2O

2C2H2 + 5O2 = 4CO2 + 2H2O

A lot of this is trial and error.  I start from left to right.  For the first reaction, you had

C2H6 + O2  =  CO2  +  H2O

You have 2 carbons on the left so you can try adding a 2 in front of the carbons on the right and keep going from there.  Now you have

C2H6  +  O2  =  2CO2  + H2O

You now have 2 C on each side.  C2 is the same as 2C.

The next step is to look at the hydrogens on the left.  You have 6 so you need 6 on the right.  You can add a 3 in front of the H on the right.  The H already has a subscript of a 2 so having the 3 in front will now make it 6H.  You now have this

C2H6  +  O2  =  2CO2  +  3H2O

So your C are equal (2 on each side) and your H is equal (6 on each side).  All that is left is the O.  So far you have 2 on the left but you have 7 on the right (4 from the 2CO2 and 3 more from 3H2O).  Since you have an even number on one side and an odd number on the other, there is no way to make this work.  So it's time to start over.

My next step was to put a 2 in front of C2H6 and keep going as we did above.  Start with this

2C2H6   + O2  = CO2  + H2O

You have 4 C on the left.  So you need to add a 4 in front of CO2 to get 4 C on the right.  Now you have this

2C2H6  +  O2  =  4CO2  +  H2O

Now on the left you have 12 H (2 in front times the 6 subscript on the H).  You need to have 12 on the right.  We can get that by putting a 6 in front of the H2O ( again the 6 in front times the 2 subscript of the H is 12 total).  You have this now

2C2H6  +  O2  = 4CO2  +  6H2O

You now have 4 C on both sides, 12 H on both sides.  All that is left now is the O.

On the right side, you have a total of 14 ( 8 from the 4CO2 and 6 from 6H2O).  So you'd need to add 7 in front of the O2 to make it 14.

You're done.  4 C on both sides, 12 H on both sides and 14 O on both sides.

The second equation I just did the same process.

6 0
3 years ago
Equivalent mass of AL2O3​
Maru [420]

Answer:

Mass = 102g

Explanation:

Given

Compound: Al_2O_3

Required

Determine the equivalent mass

In the above compound, we have: 2 Al and 3 (O)

The atomic mass of Aluminium is:

Al = 27

The atomic mass of Oxygen is:

(O) = 16

So, the equivalent mass is:

Mass = 2 * 27 + 3 * 16

Mass = 54 + 48

Mass = 102g

5 0
3 years ago
When any salt is dissolved in water, the ionic bonds between the atoms must be broken in order for the salt to dissolve. The ene
Ksju [112]

Answer:

C released to the environment

Explanation:

I just did it in class

5 0
3 years ago
Explain why NO is more soluble in water than either N2 or O2. Fill in the following blanks.
jek_recluse [69]

Answer:

a. Larger molecular size

b. Ability to react with water

c. Polarity

d. Shorter bond length

Explanation:

Ethene is a larger molecule than oxygen and nitrogen hence it is more soluble than the both other gases .

SO2 dissolves readily in water to yield an acid solution. It is an acid anhydride.

Nitric oxide is a polar compound. It remains very much polar while nitrogen and oxygen are non polar.

Nitrogen is sp hybrized, this leads to a very short bond and does not easily interact with oxygen and nitrogen

3 0
3 years ago
Calculate the concentration of IO−3 in a 1.65 mM Pb(NO3)2 solution saturated with Pb(IO3)2 . The Ksp of Pb(IO3)2 is 2.5×10−13 .
lara31 [8.8K]

Answer:

See explaination

Explanation:

1) Pb(NO3)2 => Pb2+ + 2 NO3-

[Pb2+] = [Pb(NO3)2] = 7.56 mM = 7.56 x 10-3 M

Pb(IO3)2 <=> Pb2+ + 2 IO3-

Ksp = [Pb2+][IO3-]2 = 2.5 x 10-13

7.56 x 10-3 x [IO3-]2 = 2.5 x 10-13

[IO3-] = 5.75 x 10-6 M ≈ 5.8 x 10-6 M

(2) [Pb2+] = 1.7 x 10-6 M

Ksp = [Pb2+][IO3-]2 = 2.5 x 10-13

1.7 x 10-6 x [IO3-]2 = 2.5 x 10-13

[IO3-] = 3.83 x 10-4 M

[IO3-]from Pb(IO3)2 = 2 x [Pb2+]

= 2 x 1.7 x 10-6 = 3.4 x 10-6 M

[IO3-]from NaIO3 = [IO3-] - [IO3-]from Pb(IO3)2

= 3.83 x 10-4 - 3.4 x 10-6

= 3.80 x 10-4 M

NaIO3 => Na+ + IO3-

[NaIO3] = [IO3-]from NaIO3

= 3.80 x 10-4 M ≈ 3.8 x 10-4 M

7 0
4 years ago
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