Question

These questions are based on the following balanced equation:

2H2​+O2​ 2H2​O

If you start with 6.933 g of H₂ and 3.525 g of O₂, find the following: 

With excess O₂, what mass of H₂O would be produced by the H₂?

With excess H₂, what mass of H₂O would be produced by the O₂?

Write the chemical formula for the limiting reactant:

Answer 1

Moles of $H_2 = \frac { 6.933}{2}=3.47 moles$

Moles of $O_2 = \frac {3.525}{16}=0.22moles$

With excess $O_2$ Mass of $H_2O =$ $2×0.22×18=7.92g$

With excess $H_2$ Mass of $H_2O =3.47×18=62.46$

$H_2$

Is the limiting reactant