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Zina [86]
3 years ago
12

If 15.60 g of a hydrated compound is heated in an oven for several hours, its mass drops to 8.63 g. Assuming that the reduction

in mass is due to loss of water, what is the percentage of water in the original hydrate?
Chemistry
2 answers:
Ber [7]3 years ago
6 0
In order to calculate the percent of water in the hydrate, we need to know the amount of water that is lost after heating. We simply subtract the initial amount and the final amount. We do as follows:

mass of water = 15.60 g - 8.63 g = 6.97 g

Percent water = 6.97 / 15.60 x 100 = 44.68%
puteri [66]3 years ago
6 0
Mass of water = 15.60 g - 8.63 g = 6.97 g

Percent water = 6.97 / 15.60 x 100 = 44.68%
Good Luck, I hope this helps you!! :)
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A sample of sodium reacts completely with 0.213 kgkg of chlorine, forming 351 gg of sodium chloride. What mass of sodium reacted
lapo4ka [179]

Answer:

The mass of sodium reacted is 138 grams

Explanation:

Firstly, the chemical reaction should be represented with a balance chemical equation. The chemical equation can be written as follows.

Na = sodium

Cl2 = chlorine gas

Na + Cl2 → NaCl

The balanced equation is

2Na(s) + Cl2(g) → 2NaCl(aq)

Atomic mass of sodium = 23 grams per mol

Atomic mass of chlorine = 35.5 grams per mol

From the chemical equation the molar mass are as follows

Sodium = 2 × 23 = 46 grams

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Sodium chloride =  2 × 23 + 2 × 35.5 = 117 grams

if 46 grams sodium reacted to produce 117 grams of sodium chloride

? grams of sodium react to produce 351 grams sodium chloride

grams of sodium reacted = (351 × 46)/117

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grams of sodium reacted = 138  grams

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