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Oksanka [162]
3 years ago
14

Given the reaction _K(s) +_ Cl2(g) → _KCl(s) what is the amount of K, in grams, needed to completely react with 2 moles of Cl2(g

)?
Chemistry
1 answer:
damaskus [11]3 years ago
3 0

Answer:

156.4g K

Explanation:

I'm not sure if it is correct but I think it should be this

What do we know so far?: 2K + 1Cl2 -> 2KCl, 2 mol of Cl2

What are we looking for?: #g of K

What is the ratio of K to Cl2?: 2:1

Set up equation: 2molCl2 x \frac{2mol K}{1 mol Cl2}

Cancel unwanted units: 2 x \frac{2mol K}{1}

Answer we got: 2 x 2mol K = 4mol K

Converting moles to grams: 4 x 39.1 (molar mass of K) = 156.4g K

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Answer:

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Explanation:

Given :

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m_{evaporating\ dish}+m_{Second\ heated\ sample}=13.266\ g

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m_{Hydrate\ sample}=25.637-m_{evaporating\ dish}\ g=25.637-1.135\ g=24.502\ g

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m_{Second\ heated\ sample}=m_{salt\ anhydrous}=13.266-m_{evaporating\ dish}\ g=13.266-1.135\ g=12.131\ g

Mass of water:

m_{water}=m_{Hydrate\ sample}-m_{salt\ anhydrous}=24.502-12.131\ g=12.371\ g

m_{water}=12.371\ g

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