How many grams of oxygen are required to produce 55.6 g water vapor?

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How many grams of oxygen are required to produce 55.6 g water vapor?

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Answer 1 · 2016-07-10T18:34:12

Approx. $50 \cdot g$ $50*g$ dioxygen are necessary.

Explanation:

$H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (g)$ $H_2(g) + 1/2O_2(g) rarr H_2O(g)$

$Moles of water$ $"Moles of water"$ $=$ $=$ $\frac{55.6 \cdot g}{18.01 \cdot g \cdot m o l^{- 1}}$ $(55.6*g)/(18.01*g*mol^-1)$ $=$ $=$ $3.09 \cdot m o l$ $3.09*mol$

If there are $3.09 \cdot m o l$ $3.09*mol$ of water, then clearly, there were $1.54 \cdot m o l$ $1.54*mol$ $O_{2} (g)$ $O_2(g)$ as a reactant.

And thus $mass of oxygen$ $"mass of oxygen"$ $=$ $=$ $1.54 \cdot m o l \times 32.00 \cdot g \cdot m o l^{- 1}$ $1.54*molxx32.00*g*mol^-1$ .

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How many grams of oxygen are required to produce 55.6 g water vapor?

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