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3 years ago

6

Blank c2h4 + blank o2 → blank co2 + blank h2o how many moles of o2 are in the chemical equation when balanced using the lowest w

hole numbers

1 answer:

umka21 [38]3 years ago

8 0

The reaction is<span>
C</span>₂H₄ + O₂ → CO₂ + H₂O<span>

To balance the equation, both side have same number of elements. Here,</span>

In left hand side has in right hand side has

4 H atoms 2 H atoms

2 C atoms 1 C atom

<span> 2 O atoms 3 O atoms

First, we have to balance number of C atoms and number of H atoms in both side.
To balance C atoms, '2' should be added before CO</span>₂ and to balance H atoms, '2' should be added before H₂<span>O.
Then number of oxygen atoms is </span>2 x 2 + 2 = 6 in right hand side. So, 3 should be added before O₂<span> in left hand side.
After balancing the equation should be,</span>

C₂H₄<span> + 3O</span>₂<span> → 2CO</span>₂<span> + 2H</span>₂O

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which method would be best for separating the compounds of a mixture that is made from two diffrent liquids

kobusy [5.1K]

Answer:

fractional distillation since ot depends on the different liquids to have different boiling points

5 0

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.

<u>Answer:</u> The empirical formula for the given compound is $CaCN_2$

<u>Explanation:</u>

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

<u>For calculating the mass of carbon:</u>

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.

<u>For calculating the mass of nitrogen:</u>

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.

<u>For calculating the mass of calcium:</u>

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:

<u>Step 1:</u> 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 = $\frac{\text{Given mass of Nitrogen}}{\text{Molar mass of Nitrogen}}=\frac{1.193g}{14g/mole}=0.0852moles$

<u>Step 2:</u> 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 = $\frac{0.0852}{0.0426}=2$

<u>Step 3:</u> 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 $CaCN_2$

3 0

3 years ago

Silver sulfadiazine burn-treating cream creates a barrier against bacterial invasion and releases antimicrobial agents directly

trapecia [35]

Answer : The mass of silver sulfadiazine produced can be, 71.35 grams.

Solution : Given,

Mass of $Ag_2O$ = 25.0 g

Mass of $C_{10}H_{10}N_4SO_2$ = 50.0 g

Molar mass of $Ag_2O$ = 231.7 g/mole

Molar mass of $C_{10}H_{10}N_4SO_2$ = 250.3 g/mole

Molar mass of $AgC_{10}H_{9}N_4SO_2$ = 357.1 g/mole

First we have to calculate the moles of $Ag_2O$ and $C_{10}H_{10}N_4SO_2$ .

$\text{ Moles of }Ag_2O=\frac{\text{ Mass of }Ag_2O}{\text{ Molar mass of }Ag_2O}=\frac{25.0g}{231.7g/mole}=0.1079moles$

$\text{ Moles of }C_{10}H_{10}N_4SO_2=\frac{\text{ Mass of }C_{10}H_{10}N_4SO_2}{\text{ Molar mass of }C_{10}H_{10}N_4SO_2}=\frac{50.0g}{250.3g/mole}=0.1998moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$Ag_2O(s)+2C_{10}H_{10}N_4SO_2(s)\rightarrow 2AgC_{10}H_9N_4SO_2(s)+H_2O(l)$

From the balanced reaction we conclude that

As, 2 mole of $C_{10}H_{10}N_4SO_2$ react with 1 mole of $Ag_2O$

So, 0.1998 moles of $C_{10}H_{10}N_4SO_2$ react with $\frac{0.1998}{2}=0.0999$ moles of $Ag_2O$

From this we conclude that, $Ag_2O$ is an excess reagent because the given moles are greater than the required moles and $C_{10}H_{10}N_4SO_2$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $AgC_{10}H_9N_4SO_2$

From the reaction, we conclude that

As, 2 mole of $C_{10}H_{10}N_4SO_2$ react with 2 mole of $AgC_{10}H_9N_4SO_2$

So, 0.1998 mole of $C_{10}H_{10}N_4SO_2$ react with 0.1998 mole of $AgC_{10}H_9N_4SO_2$

Now we have to calculate the mass of $AgC_{10}H_9N_4SO_2$

$\text{ Mass of }AgC_{10}H_9N_4SO_2=\text{ Moles of }AgC_{10}H_9N_4SO_2\times \text{ Molar mass of }AgC_{10}H_9N_4SO_2$

$\text{ Mass of }AgC_{10}H_9N_4SO_2=(0.1998moles)\times (357.1g/mole)=71.35g$

Therefore, the mass of silver sulfadiazine produced can be, 71.35 grams.

8 0

3 years ago

A molecule of hydrogen gas is made up of pairs of hydrogen atoms in which each atom shares its electron to form a(n) ____.

andreyandreev [35.5K]

Answer:

covalent bond

Explanation:

a covalent bond forms when electrons are shared between two nonmetals

plz mark as the brainliest

6 0

2 years ago

A 34.57 mL sample of an unknown phosphoric acid solution is titrated with a 0.127 M sodium hydroxide solution. The equivalence p

Bas_tet [7]

<u>Answer:</u> The concentration of unknown phosphoric acid solution is 0.034 M

<u>Explanation:</u>

To calculate the concentration of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $H_3PO_4$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=3\\M_1=?M\\V_1=34.57mL\\n_2=1\\M_2=0.127M\\V_2=28.2mL$

Putting values in above equation, we get:

$3\times M_1\times 34.57=1\times 0.127\times 28.2\\\\M_1=\frac{1\times 0.127\times 28.2}{3\times 34.57}=0.034M$

Hence, the concentration of unknown phosphoric acid solution is 0.034 M

7 0

3 years ago