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natima [27]
4 years ago
14

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

Chemistry
2 answers:
grin007 [14]4 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]4 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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Moles of cu = mass/Ar
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Vol of cu = ans/1 × 24dm3
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The Handbook of Chemistry and Physics gives solubilities of the following compounds in grams per 100 mL water. Because these com
Elanso [62]

Answer:

(a) Ksp=4.50x10^{-7}

(b) Ksp=1.55x10^{-6}

(c) Ksp=2.27x10^{-12}

(d) Ksp=1.05x10^{-22}

Explanation:

Hello,

In this case, given the solubility of each salt, we can compute their molar solubilities by using the molar masses. Afterwards, by using the mole ratio between ions, we can compute the concentration of each dissolved and therefore the solubility product:

(a) BaSeO_4(s)\rightleftharpoons Ba^{2+}(aq)+SeO_4^{2-}(aq)

Molar\ solubility=\frac{0.0188g}{100mL} *\frac{1mol}{280.3g}*\frac{1000mL}{1L}=6.7x10^{-4}\frac{mol}{L}

In such a way, as barium and selenate ions are in 1:1 molar ratio, they have the same concentration, for which the solubility product turns out:

Ksp=[Ba^{2+}][SeO_4^{2-}]=(6.7x10^{-4}\frac{mol}{L}   )^2\\\\Ksp=4.50x10^{-7}

(B) Ba(BrO_3)_2(s)\rightleftharpoons Ba^{2+}(aq)+2BrO_3^{-}(aq)

Molar\ solubility=\frac{0.30g}{100mL} *\frac{1mol}{411.15g}*\frac{1000mL}{1L}=7.30x10^{-3}\frac{mol}{L}

In such a way, as barium and bromate ions are in 1:2 molar ratio, bromate ions have twice the concentration of barium ions, for which the solubility product turns out:

Ksp=[Ba^{2+}][BrO_3^-]^2=(7.30x10^{-3}\frac{mol}{L})(3.65x10^{-3}\frac{mol}{L})^2\\\\Ksp=1.55x10^{-6}

(C) NH_4MgAsO_4(s)\rightleftharpoons NH_4^+(aq)+Mg^{2+}(aq)+AsO_4^{3-}(aq)

Molar\ solubility=\frac{0.038g}{100mL} *\frac{1mol}{289.35g}*\frac{1000mL}{1L}=1.31x10^{-4}\frac{mol}{L}

In such a way, as ammonium, magnesium and arsenate ions are in 1:1:1 molar ratio, they have the same concentrations, for which the solubility product turns out:

Ksp=[NH_4^+][Mg^{2+}][AsO_4^{3-}]^2=(1.31x10^{-4}\frac{mol}{L})^3\\\\Ksp=2.27x10^{-12}

(D) La_2(MoOs)_3(s)\rightleftharpoons 2La^{3+}(aq)+3MoOs^{2-}(aq)

Molar\ solubility=\frac{0.00179g}{100mL} *\frac{1mol}{1136.38g}*\frac{1000mL}{1L}=1.58x10^{-5}\frac{mol}{L}

In such a way, as the involved ions are in 2:3 molar ratio, La ion is twice the molar solubility and MoOs ion is three times it, for which the solubility product turns out:

Ksp=[La^{3+}]^2[MoOs^{-2}]^3=(2*1.58x10^{-5}\frac{mol}{L})^2(3*1.58x10^{-5}\frac{mol}{L})^3\\\\Ksp=1.05x10^{-22}

Best regards.

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