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3 years ago

g How many grams of beryllium are needed to produce "11.5" g of hydrogen gas?Be(s) + 2H 2 O(l) → Be(OH) 2 (aq) + H 2 (g)

Chemistry

1 answer:

slega [8]3 years ago

6 0

Answer:

$m_{Be}=51.3gBe$

Explanation:

Hello,

In this case, with the given chemical reaction:

$Be(s) + 2H_2 O(l) \rightarrow Be(OH)_ 2 (aq) + H _2 (g)$

We can notice a 1:1 mole ratio between hydrogen gas and beryllium metal, therefore, beryllium requirement is computed as shown below:

$m_{Be}=11.5gH_2*\frac{1molH_2}{2.02gH_2} *\frac{1molBe}{1molH_2} *\frac{9.01gBe}{1molBe}\\ \\m_{Be}=51.3gBe$

Regards.

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<h3>What is Empirical Formula ?</h3>

Empirical formula is the simplest whole number ratio of atoms present in given compound.

Element % Atomic mass Relative no. of atoms Simplest whole ratio

C 40.6 12 $\frac{40.6}{12}$ = 3.3 $\frac{3.3}{3.3} = 1$

H 5.1 1 $\frac{5.1}{1}$ = 5.1 $\frac{5.1}{3.1} = 2$

O 54.2 16 $\frac{54.2}{16}$ = 3.3 $\frac{3.3}{3.3} = 1$

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<h3>How to find the Molecular formula of compound ?</h3>

Molecular formula = Empirical formula × n

$n = \frac{\text{Molecular weight}}{\text{Empirical Formula weight}}$

$= \frac{118.084}{30}$

= 4

Molecular formula = n × Empirical formula

= 4 (C₁H₂O₁)

Thus from the above conclusion we can say that The Empirical formula of compound is C₁H₂O₁. The Molecular Formula of the compound is 4 (C₁H₂O₁).

Learn more about the Empirical Formula here: brainly.com/question/1603500

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Explanation:

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