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3 years ago

C3H8 + xO2 > 3CO2 + yH2O which value of x and y balance the equations

Chemistry

1 answer:

Gelneren [198K]3 years ago

8 0

Answer:

x = 5 and y =4

Explanation:

An alkane is a hydrocarbon of general molecular formula CnH2n+2. The general formula for balancing the combustion equation of any alkane is;

CnH2n+2 + 3n+1/2O2 ------> nCO2 + (n+1) H2O

Where n is the number of carbon atoms in the alkane.

In C3H8 , n=3 so;

C3H2(3)+2 + 3(3)+1/2O2 ------> 3CO2 + (3+1) H2O

Then;

C3H8 + 5O2 ------> 3CO2 + 4H2O

Therefore;

x = 5 and y =4

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Answer : The final volume of gas will be, 26.3 mL

Explanation :

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

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$T_1$ = initial temperature of gas = $22.0^oC=273+22.0=295K$

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$\frac{0.974 atm\times 27.5 mL}{295K}=\frac{0.993 atm\times V_2}{288K}$

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3 years ago

Read 2 more answers

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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3 years ago

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finlep [7]

Answer:

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Answer : Option (A) Accelerator 2 model has the lowest percentage of energy lost as waste.

Solution : Given,

For Accelerator 1 model,

Input energy = 2078.3 J

Wasted energy = 663.1 J

Output energy = 1415.2 J

For Accelerator 2 model,

Input energy = 7690.0 J

Wasted energy = 2337.5 J

Output energy = 5353.5 J

For Accelerator 3 model,

Input energy = 4061.9 J

Wasted energy = 2259.6 J

Output energy = 1802.3 J

Formula used for lowest percentage of energy lost as waste is:

% energy lost as waste = (Total energy wasted / Total input energy ) × 100

For Accelerator 1 model,

% energy lost as waste = $\frac{663.1}{2078.3}\times100$ = 31.90%

For Accelerator 2 model,

% energy lost as waste = $\frac{2337.5}{7690.0}\times100$ = 30.39%

For Accelerator 3 model,

% energy lost as waste = $\frac{2259.6}{4061.9}\times100$ = 55.62%

So, we conclude that the Accelerator 2 model has the lowest percentage of energy lost as waste.

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3 years ago