All categories

3 years ago

Balance the following reaction. CH4+O2==>CO2+H20

Chemistry

2 answers:

Radda [10]3 years ago

5 0

CH4 + 2O2 --> CO2 + 2H2O

tatiyna3 years ago

3 0

Answer:

$\huge \boxed{\mathrm{CH_4+ 2O_2 \Rightarrow CO_2 +2 H_2O}}$

$\rule[225]{225}{2}$

Explanation:

$\sf CH_4+ O_2 \Rightarrow CO_2 + H_2O$

Balancing the Hydrogen atoms on the right side,

$\sf CH_4+ O_2 \Rightarrow CO_2 +2 H_2O$

Balancing the Oxygen atoms on the left side,

$\sf CH_4+ 2O_2 \Rightarrow CO_2 +2 H_2O$

$\rule[225]{225}{2}$

You might be interested in

I really need some help and explaining with the Heat of Combustion Lab.

Nataly_w [17]

Combustion means the elements or compound can be burned, but burning which is a chemical process requires oxygen; combustion reaction typically takes place in the presence of air. The combustion of methane is as follows: CH4(g) + 2 O2(g) -> CO2(g)+ 2 H2O(g) + energy One mole of gaseous methane reacts with two oxygen molecules to form a carbon dioxide molecule, and two water molecules which is given off as water vapor. The reaction involves the release of heat.

6 0

3 years ago

In the following reaction, H3PO4 (aq) + H2O (l) ⇄ H2PO4– (aq) + H3O+ (aq) what happens when more H2PO4– (aq) is added to the sol

kirza4 [7]

Answer:

The concentration of H₃PO₄ will increase.

Explanation:

H₃PO₄(aq) + H₂O(l) ⇄ H₂PO₄⁻(aq) + H₃O⁺(aq)

According to Le Châtelier's Principle, when we apply a stress to a system at equilibrium, the system will respond in a way that tends to relieve the stress.

If we add more H₂PO₄⁻, the position of equilibrium will move to the left to get rid of the added H₂PO₄⁻.

The concentration of H₃PO₄ will increase.

7 0

3 years ago

A gaseous compound is 30.4 % N and 69.6% OF. A 5.25 g sample of the gas occupies a volume of 1.00 L and exerts a pressure of 958

Harman [31]

Answer:

The molecular formula = N2O4

Explanation:

Step 1: Data given

A gaseous compound is 30.4 % N and 69.6%

Mass of the compound = 5.25 grams

Volume of the gas = 1.00 L

Pressure of the gas = 958 mmHg = 1.26 atm

Temperature of the gas = -4 °C = 273 -4°C = 269 Kelvin

Molar mass of N = 14 g/mol

Molar mass of O = 16 g/mol

Step 2: Calculate mass of N

Mass of Nitrogen = 5.25 grams * 0.304 = 1.596 grams

Step 3: Calculate mass of O

Mass of Oxygen = 5.25 grams * 0.696 = 3.654 grams

Step 4: Calculate number of moles N

Number of moles N = Mass of N/ Molar mass of N

Moles of N = 1.596 grams / 14g/mol

Moles of N = 0.114 moles N

Step 5: Calculate moles of O

Moles O = 3.654 grams / 16 g/mol

Moles 0 = 0.2884 moles

Step 6: Calculate empirical formule

We calculate the empirical formule by dividing number of moles by the smallest number of mol

N : 0.114 / 0.114 = 1

O: 0.2284 / 0.114 = 2

Empirical formule = NO2

Step 7: Calculate number of moles of 5.25 g sample via gas law:

p*V = nRT

⇒ with p = the pressure = 1.26 atm

⇒ with v = 1.00 L

⇒ with n = the number of moles = TO BE DETERMINED

⇒ with R = the gas constant = 0.08206 L*atm/ K*mol

⇒ with T = the temperature = 269 K

number of moles n = (p*V)/(R*T)

n = (1.26*1L)/(0.08206*269)

n = 0.057 mol

Step 8: Calculate molar mass of the compound

This means 5.25 grams of the gas = 0.057 moles

So 1 mol of the compound has a molar mass of: 5.25 / 0.057 = 92.11 g/mol

Step 9: Calculate molar mass of the empirical formula NO2

N = 14 g/mol

O = 16 g/mol

NO2 = 14 + 16 + 16 = 46 g/mol

The empirical formule NO2 has a molar mass of 46 g/mol

Step 10: Calculate molecular formula

92.11 / 46 = 2

This means the empirical formula should be multiplied by 2

2*(NO2) = N2O4

The molecular formula = N2O4

8 0

3 years ago

In a reaction vessel, 17.6 g of solid chromium(III) oxide, Cr2O3, was allowed to react with excess carbon tetrachloride in the f

notsponge [240]

Answer:

72.53% is the yield of CrCl3

Explanation:

Given

Reaction:

Cr2O3(s) + 3 CCl4(l) → 2 CrCl3(s) + 3 COCl2(aq)

CCl4 is in excess and 17.6g Cr2O3 present

The reaction yields 26.6g of CrCl3

To Find:

% yields of the reaction

Also given

Molar mass of CrCl3 = 158.35g/mol

Molar mass of Cr2O3 = 152.00 g/mol

By the stoichiometry of the reaction

1 mole of Cr2O3 gives 2 moles of CrCl3

1 x1 52 g of Cr2O3 gives 2x 158.35 g of CrCl3

= 1 52 g of Cr2O3 gives 316.70 g of CrCl3

17.6 g of Cr2O3 gives (17.6÷152) × 316.70 g CrCl3

= 36.67 g CrCl3

but actual yield is only 26.6g

so % yield is (26.6 ÷÷ 36.67) × 100

= 72.53% is the yield of CrCl3

8 0

4 years ago

Calculate the new volume of 1.23 mL of a gas at 32 C is subjected to drop in temperature of 20 degrees Celsius

kotykmax [81]

Answer:

1,15mL = V₂

Explanation:

Based on Charle's law the volume is directely proportional to the absolute temperature in a gas under constant pressure. The equation is:

V₁T₂ = V₂T₁

Where V is volume and T absolute temperature of a gas where 1 is initial state and 2, final state.

The V₁ is 1.23mL

T₁ = 32°C + 273.15 = 305.15K

T₂ = T₁ - 20°C = 285.15K

Replacing:

1.23mL*285.15K = V₂*305.15K

8 0

3 years ago