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3 years ago

How many moles of hexane must burn to form 18.4 mol of carbon dioxide

Chemistry

2 answers:

romanna [79]3 years ago

6 0

Answer : The number of moles of hexane is, 3.06 moles

Explanation : Given,

Moles of carbon dioxide = 18.4 mole

The balanced chemical reaction will be,

$2C_6H_{14}+19O_2\rightarrow 12CO_2+14H_2O$

From the balanced chemical reaction, we conclude that 2 moles of hexane react with 19 moles of oxygen gas to give 12 moles of carbon dioxide gas and 14 moles of water as a product.

As, 12 moles of carbon dioxide obtained from 2 moles of hexane

So, 18.4 moles of carbon dioxide obtained from $\frac{2}{12}\times 18.4=3.06$ moles of hexane

Therefore, the number of moles of hexane is, 3.06 moles

Talja [164]3 years ago

5 0

He combustion reaction of hexane:

2 C6H14 + 19 O2 = 12 CO2 + 14 H2O

2 mol C6H14 ---------> 12 mol CO2
x mol C6H14 ---------> 18.4 mol CO2

12 * x = 18.4 * 2

12 x = 36.8

x = 36.8 / 12

x = 3.067 moles of CO2

hope this helps!.

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2 years ago

Burning a compound of calcium, carbon, and nitrogen in oxygen in a combustion train generates calcium oxide , carbon dioxide , n

mylen [45]

The question is incomplete, here is the complete question:

Burning a compound of calcium, carbon, and nitrogen in oxygen in a combustion train generates calcium oxide (CaO), carbon dioxide $(CO_2)$ , nitrogen dioxide $(NO_2)$ , and no other substances. A small sample gives 2.389 g CaO, 1.876 g $CO_2$ , and 3.921 g $NO_2$ Determine the empirical formula of the compound.

Answer: The empirical formula for the given compound is $CaCN_2$

Explanation:

The chemical equation for the combustion of compound having calcium, carbon and nitrogen follows:

$Ca_xC_yN_z+O_2\rightarrow CaO+CO_2+NO_2$

where, 'x', 'y' and 'z' are the subscripts of calcium, carbon and nitrogen respectively.

We are given:

Mass of CaO = 2.389 g

Mass of $CO_2=1.876g$

Mass of $NO_2=3.921g$

We know that:

Molar mass of calcium oxide = 56 g/mol

Molar mass of carbon dioxide = 44 g/mol

Molar mass of nitrogen dioxide = 46 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 1.876 g of carbon dioxide, $\frac{12}{44}\times 1.876=0.5116g$ of carbon will be contained.

For calculating the mass of nitrogen:

In 46 g of nitrogen dioxide, 14 g of nitrogen is contained.

So, in 3.921 g of nitrogen dioxide, $\frac{14}{46}\times 3.921=1.193g$ of nitrogen will be contained.

For calculating the mass of calcium:

In 56 g of calcium oxide, 40 g of calcium is contained.

So, in 2.389 g of calcium oxide, $\frac{40}{56}\times 2.389=1.706g$ of calcium will be contained.

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Calcium = $\frac{\text{Given mass of Calcium}}{\text{Molar mass of Calcium}}=\frac{1.706g}{40g/mole}=0.0426moles$

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.5116g}{12g/mole}=0.0426moles$

Moles of Nitrogen = $\frac{\text{Given mass of Nitrogen}}{\text{Molar mass of Nitrogen}}=\frac{1.193g}{14g/mole}=0.0852moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.0426 moles.

For Calcium = $\frac{0.0426}{0.0426}=1$

For Carbon = $\frac{0.0426}{0.0426}=1$

For Nitrogen = $\frac{0.0852}{0.0426}=2$

Step 3: Taking the mole ratio as their subscripts.

The ratio of Ca : C : N = 1 : 1 : 2

Hence, the empirical formula for the given compound is $CaCN_2$

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2 years ago

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densk [106]

Answer:

Close to the calculated endpoint of a titration - Partially open

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