For the reaction C + 2H2 → CH4, how many grams of hydrogen are required to produce 13.1 moles of methane, CH4 ?

When we write the formula for an ionic compound, we are merely indicate the "molecules" in the compound with that formula. expla

belka [17]

When writing an ionic compound formula, a "molecular" form is used. The formula is made with allowance for ion charges.
For example,
Ca²⁺ and NO₃⁻ ⇒ Ca(NO₃)₂
Al³⁺ and SO₄²⁻ ⇒ Al₂(SO₄)₃

3 0

3 years ago

How do scientists use sonar to study Earth’s oceans?

adoni [48]

A to map the ocean floor

7 0

3 years ago

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Liquid hexane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 6.9 g of hexane is mi

MrRa [10]

Answer:

There, there are no leftover for C6H14 instead, an additional mass of 4g of C6H14 is needed to completely react with 38.4g of O2.

Explanation:

Step 1:

We'll begin by writing the balanced equation for the reaction. This is shown below:

2C6H14 + 19O2 —> 12CO2 + 14H2O

Step 2:

Let us calculate the masses of C6H14 and O2 that reacted from the balanced equation. This is illustrated below:

2C6H14 + 19O2 —> 12CO2 + 14H2O

Molar Mass of C6H14 = (12x6) + (14x1) = 72 + 14 = 86g/mol

Mass of C6H14 from the balanced equation = 2 x 86 = 172g

Molar Mass of O2 = 16x2 =32g/mol

Mass of O2 from the balanced equation = 19 x 32 = 608g

From the balanced equation above, 172g of C6H14 reacted with 608g of O2.

Step 3.

Now, let us determine the mass of C6H14 that will react with 38.4 g of oxygen. This is illustrated below:

From the balanced equation above, 172g of C6H14 reacted with 608g of O2.

Therefore, Xg of C6H14 will react with 38.4g of O2 i.e

Xg of C6H14 = (172 x 38.4) /608

Xg of C6H14 = 10.9g

From the calculations made above, we can see clearly that the mass of C6H14 is limited as the reaction requires 10.9g of C6H14 and only 6.9g was given. There, there are no leftover for C6H14 instead, an additional mass ( 10.9 - 6.9 = 4g) of 4g of C6H14 is needed to completely react with 38.4g of O2.

5 0

3 years ago

A 0.1510 gram sample of a hydrocarbon produces 0.5008 gram CO2 and 0.1282 gram H2O in combustion analysis. Its

Over [174]

In a combustion of a hydrocarbon compound, 2 reactions are happening per element:

C + O₂ → CO₂
2 H + 1/2 O₂ → H₂O

Thus, we can determine the amount of C and H from the masses of CO₂ and H₂O produced, respectively.

1.) Compute for the amount of C in the compound. The data you need to know are the following:
Molar mass of C = 12 g/mol
Molar mass of CO₂ = 44 g/mol
Solution:
0.5008 g CO₂*(1 mol CO₂/ 44 g)*(1 mol C/1 mol CO₂) = 0.01138 mol C
0.01138 mol C*(12 g/mol) = 0.13658 g C

Compute for the amount of H in the compound. The data you need to know are the following:
Molar mass of H = 1 g/mol
Molar mass of H₂O = 18 g/mol
Solution:
0.1282 g H₂O*(1 mol H₂O/ 18 g)*(2 mol H/1 mol H₂O) = 0.014244 mol H
0.014244 mol H*(1 g/mol) = 0.014244 g H

The percent composition of pure hydrocarbon would be:
Percent composition = (Mass of C + Mass of H)/(Mass of sample) * 100
Percent composition = (0.13658 g + 0.014244 g)/(<span>0.1510 g) * 100
</span>Percent composition = 99.88%

2. The empirical formula is determined by finding the ratio of the elements. From #1, the amounts of moles is:

Amount of C = 0.01138 mol
Amount of H = 0.014244 mol

Divide the least number between the two to each of their individual amounts:
C = 0.01138/0.01138 = 1
H = 0.014244/0.01138 = 1.25

The ratio should be a whole number. So, you multiple 4 to each of the ratios:
C = 1*4 = 4
H = 1.25*4 = 5

Thus, the empirical formula of the hydrocarbon is C₄H₅.

3. The molar mass of the empirical formula is

Molar mass = 4(12 g/mol) + 5(1 g/mol) = 53 g/mol
Divide this from the given molecular weight of 106 g/mol
106 g/mol / 53 g/mol = 2
Thus, you need to multiply 2 to the subscripts of the empirical formula.

Molecular Formula = C₈H₁₀

4 0

3 years ago

Please help!!

antiseptic1488 [7]

Answer:

Explanation:

Scientific laws or laws of science are statements, based on repeated experiments or observations, that describe or predict a range of natural phenomena.[1] The term law has diverse usage in many cases (approximate, accurate, broad, or narrow) across all fields of natural science (physics, chemistry, astronomy, geoscience, biology). Laws are developed from data and can be further developed through mathematics; in all cases they are directly or indirectly based on empirical evidence. It is generally understood that they implicitly reflect, though they do not explicitly assert, causal relationships fundamental to reality, and are discovered rather than invented.[2]

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3 years ago