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In this reaction O2 is the excess reactant and H2 is the limiting reactant, because:

n(H2)/2 < n(O2)/1 ; 1<1.55;

1 and 2 - stoichiometric coefficients of O2 and H2 respectively;

That's why the produced H2O is calculeted by the amount of H2 (the limitant reactant);

n(H2O)=n(H2)* (stoichiometric coefficients of H2O) / (stoichiometric coefficients of H2)=

=2*2/2=2 mole

Answer: 2 moles of H2O can be produced

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