Science, asked by thapaavinitika6765, 8 months ago

CH_4+O_2\Rightarrow \:CO_2+H_2O

Answers

Answered by Anonymous
1

\mathrm{Balance}\:CH_4+O_2\Rightarrow \:CO_2+H_2O:\quad CH_4+2O_2\Rightarrow \:CO_2+2H_2O

\mathrm{Balance\:reaction\:by\:inspection}

CH_4+O_2\Rightarrow \:CO_2+H_2O

C\:\mathrm{is\:balanced}

CH_4+O_2\Rightarrow \:CO_2+H_2O

\mathrm{To\:balance}\:H\:\mathrm{multiply}\:H_2O\:\mathrm{on\:the\:right\:by}\:2

CH_4+O_2\Rightarrow \:CO_2+2H_2O

\mathrm{To\:balance}\:O\:\mathrm{multiply}\:O_2\:\mathrm{on\:the\:right\:by}\:2

CH_4+2O_2\Rightarrow \:CO_2+2H_2O

Answered by Anonymous
2

Answer

\mathrm{Balance\:reaction\:by\:inspection}

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

 \sf C\:\mathrm{is\:balanced}

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

\mathrm{\sf To\:balance}\:H\:\mathrm{multiply}\:H_2O\:\mathrm{on\:the\:right\:by}\:2

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

\mathrm{To\:balance}\:O\:\mathrm{multiply}\:O_2\:\mathrm{on\:the\:right\:by}\:2

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

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