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Limiting Reagent or Reactant

How many gram...

Question

How many grams of $H_{2} O$ will be formed when $16.0$ g $H_{2}$ is mixed with $16.0$ g $O_{2}$ and allowed to react to form water?

9.0

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18.0

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32.0

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36.0

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72.0

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Solution

The correct option is B $18.0$ g
$H_{2} + \frac{1}{2} O_{2} \to H_{2} O$

1 mole of $H_{2}$ combines with a half mole of $O_{2}$ to form one mole of $H_{2} O$ .
The molecular weights of $H_{2}$ , $O_{2}$ and $H_{2} O$ are 2 g/mol, 32 g/mol and 18 g/mole respectively.
2 g (1 mole) of $H_{2}$ combines with 16 g (half mole) of $O_{2}$ to form 18 g (one mole) of $H_{2} O$ .
We have 16.0 $H_{2}$ is mixed with 16.0g $O_{2}$
Out of this, 2.0 $H_{2}$ will combine with 16.0g $O_{2}$ since oxygen mass is limited.

18 g (one mole) of $H_{2} O$ will be obtained.

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How many grams of H2O will be formed when 16.0 g H2 is mixed with 16.0 g O2 and allowed to react to form water?

How many grams of $H_{2} O$ will be formed when $16.0$ g $H_{2}$ is mixed with $16.0$ g $O_{2}$ and allowed to react to form water?