Combustion of $50 \ mL$ methane is carried out with $150 \ mL$ of air containing $21 \%$ oxygen by volume. What will be the total volume of the gaseous mixture after the reaction? (Assume water is in the gaseous state and air contains $79 \%$ nitrogen by volume).
$CH_{4(g)} + 2O_{2(g)} \to CO_{2(g)} + 2H_2O_{(g)}$

  • A
    $200$
  • B
    $110$
  • C
    $113$
  • D
    $144.5$

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