How many grams of NaOH is produced from $1.20 x 10^{2}$ $1.20 x 10^2$ grams of $N a_{2} O$ $Na_2O$ ? $N a_{2} O + H_{2} O \to 2 N a O H$ $Na_2O + H_2O -> 2NaOH$

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How many grams of NaOH is produced from $1.20 x 10^{2}$ $1.20 x 10^2$ grams of $N a_{2} O$ $Na_2O$ ? $N a_{2} O + H_{2} O \to 2 N a O H$ $Na_2O + H_2O -> 2NaOH$

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Answer 1 · 2014-07-20T04:55:21

If we begin the problem with 120. grams of $N a_{2} O$ $Na_2O$ and we are trying to find the mass of $N a O H$ $NaOH$ that can be produced, this is a gams to grams stoichiometry problem.

$g r a m s \to m o l s \to m o l s \to g r a m s$ $grams -> mols -> mols -> grams$

$120 . g N a_{2} O$ $120. g Na_2O$ x $\frac{1 m o l N a_{2} O}{62 . g N a_{2} O}$ $(1 mol Na_2O) / (62. g Na_2O)$ x $\frac{2 m o l N a O H}{1 m o l N a_{2} O}$ $(2 mol NaOH) / (1 mol Na_2O)$ x $\frac{40 . g N a O H}{1 m o l N a O H}$ $(40. g NaOH) / (1 mol NaOH)$ =

the gfm of $N a_{2} O$ $Na_2O$ is (2 x 23 + 1 x 16 = 62)
the gfm of $N a O H$ $NaOH$ is (1 x 23 + 1 x 16 + 1 x 1 = 40)

The mole ratio from the balanced chemical equation is 2 moles of $N a O H$ $NaOH$ for every mole mole of $N a_{2} O$ $Na_2O$ .

The final calculation is
120 x 2 x 40 / 62 = 154.8387

The final solution comes to $154 . g r a m s N a O H$ $154. grams NaOH$

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How many grams of NaOH is produced from $1.20 x 10^{2}$ $1.20 x 10^2$ grams of $N a_{2} O$ $Na_2O$ ? $N a_{2} O + H_{2} O \to 2 N a O H$ $Na_2O + H_2O -> 2NaOH$

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How many grams of NaOH is produced from $1.20 x 10^{2}$ $1.20 x 10^2$ grams of $N a_{2} O$ $Na_2O$ ? $N a_{2} O + H_{2} O \to 2 N a O H$ $Na_2O + H_2O -> 2NaOH$