How many grams of NaCl are produced when 80.0 grams of O2 are produced? 2 NaClO3 ---> 2 NaCl + 3 O2

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How many grams of NaCl are produced when 80.0 grams of O2 are produced? 2 NaClO3 ---> 2 NaCl + 3 O2

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Answer 1 · 2014-05-14T16:06:28

How many grams of NaCl are produced if 80.0 grams of $O_{2}$ $O_2$ are produced?

We begin with a balanced chemical equation that is provided in the question.

$2 N a C l O_{3} \to 2 N a C l + 3 O_{2}$ $2NaClO_3 -> 2NaCl + 3O_2$

Next we determine what we have and what we want.
We have 80.0 grams of Oxygen and we want grams of Sodium Chloride.

We set up a roadmap to solve the problem

$g r a m s O_{2} \to m o l O_{2} \to m o l N a C l \to g r a m s N a C l$ $grams O_2 -> mol O_2 -> mol NaCl -> grams NaCl$

We need the molar mass ( $g f m$ $gfm$ ) of $O_{2}$ $O_2$ and $N a C l$ $NaCl$ .
$O_{2}$ $O_2$ = 31.98 g/mol and $N a C l$ $NaCl$ = 58.44 g/mol.

We need the mole ratio between $O_{2}$ $O_2$ : $N a C l$ $NaCl$ 3:2. This comes from the coefficients of the balanced chemical equation.

Now we set up conversion factors following the roadmap from above.
Unit we want in the numerator, unit to cancel in the denominator.

$80.0 g O_{2} x \frac{1 m o l O_{2}}{31.98 g O_{2}} x \frac{2 m o l N a C l}{3 m o l O_{2}} x \frac{58.44 g N a C l}{1 m o l N a C l} =$ $80.0 g O_2 x (1 mol O_2)/(31.98 g O_2) x (2mol NaCl)/(3 mol O_2) x(58.44 g NaCl)/(1 mol NaCl) =$

Multiply the numerators, and divide the denominators.

The outcome is
$97.5 g N a C l$ $97.5 g NaCl$
will be produced with 80.0 grams $O_{2}$ $O_2$

I hope this was helpful.
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How many grams of NaCl are produced when 80.0 grams of O2 are produced? 2 NaClO3 ---> 2 NaCl + 3 O2

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