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natima [27]
3 years ago
14

How to equally balance CH4 + O2 → H2O + CO2 (Please show work)

Chemistry
2 answers:
grin007 [14]3 years ago
4 0

Answer:

Balanced chemical equation:

CH₄ + 2O₂  →  2H₂O + CO₂

Explanation:

Chemical equation:

CH₄ + O₂  →  H₂O + CO₂

Balanced chemical equation:

CH₄ + 2O₂  →  2H₂O + CO₂

Step one:

CH₄ + O₂  →  H₂O + CO₂

C  = 1                          C = 1

H  =  4                        H = 2

O  = 2                         O = 3

Step 2:

CH₄ + 2O₂  →  H₂O + CO₂

C  = 1                          C = 1

H  =  4                        H = 2

O  = 4                         O = 3

Step 3:

CH₄ + 2O₂  →  2H₂O + CO₂

C  = 1                          C = 1

H  =  4                        H = 4

O  = 4                         O = 4

dimaraw [331]3 years ago
4 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}

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Three.
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8 0
3 years ago
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Which of the following pairs of compounds have the same empirical formula?
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2 years ago
A 5.000 g mixture contains strontium nitrate and potassium bromide. Excess lead(II) nitrate solution is added to precipitate out
scZoUnD [109]

<u>Answer:</u> The mass percent of potassium bromide in the mixture is 9.996%

<u>Explanation:</u>

  • To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

<u>For lead (II) bromide:</u>

Given mass of lead (II) bromide = 0.7822 g

Molar mass of lead (II) bromide = 367 g/mol

Putting values in equation 1, we get:

\text{Moles of lead (II) bromide}=\frac{0.7822g}{367g/mol}=0.0021mol

  • The chemical equation for the reaction of lead (II) nitrate and potassium bromide follows:

2KBr+Pb(NO_3)_2\rightarrow PbBr_2+2KNO_3

By Stoichiometry of the reaction:

1 mole of lead (II) bromide is produced from 2 moles of potassium bromide

So, 0.0021 moles of lead (II) bromide will be produced from = \frac{2}{1}\times 0.0021=0.0042mol of potassium bromide

  • Now, calculating the mass of potassium bromide by using equation 1, we get:

Molar mass of KBr = 119 g/mol

Moles of KBr = 0.0042 moles

Putting values in equation 1, we get:

0.0042mol=\frac{\text{Mass of KBr}}{119g/mol}\\\\\text{Mass of KBr}=0.4998g

  • To calculate the percentage composition of KBr in the mixture, we use the equation:

\%\text{ composition of KBr}=\frac{\text{Mass of KBr}}{\text{Mass of mixture}}\times 100

Mass of mixture = 5.000 g

Mass of KBr = 0.4998 g

Putting values in above equation, we get:

\%\text{ composition of KBr}=\frac{0.4998g}{5.000g}\times 100=9.996\%

Hence, the percent by mass of KBr in the mixture is 9.996 %

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