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mixer [17]
3 years ago
8

You are given 3.0 grams of solid sodium to react to pure water which has a molarity of 55.6 M. How many milligrams of H2 can be

produced if you react 6 mL of the water with 3.0 grams of solid sodium.
Na (s) + H2O ------> NaOH (aq) + H2 (g)
Chemistry
1 answer:
serious [3.7K]3 years ago
4 0

Answer:

= 15.51 mL

Explanation:

Here's is the reaction:

2HgO(s) ⇒ 2 Hg(s)+O₂(g)

In this reaction 2mol HgO =  1mol O₂

The molecular weight of HgO = 216.59g

so, 3.0g HgO = 3.0g x 1.00molHgO/216.59gHgO

= 0.0138511 molHgO

The amount of Oxygen follows:

0.0138511 molHgOx1/2= 0.00692555 mol O₂

Now, volume of 1 any gas = 22400mL

so, 0.00692555 mol O₂ x22400mLO₂/1mol O₂

= 15.513232mL O₂

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Answer:

V₁  = 96.2 mL

Explanation:

Given data:

Initial volume of NH₄OH required = ?

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Solution:

Formula:

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Now we will put the values.

15.6 M ×V₁  =  3.00 M×500.0 mL

15.6  M ×V₁  = 1500 M.mL

V₁  = 1500 M.mL /15.6 M

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