Question

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)

Answer 1

`m_(Be)=51.3gBe`

Step-by-step 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*(1molH_2)/(2.02gH_2) *(1molBe)/(1molH_2) *(9.01gBe)/(1molBe)\\ \\m_(Be)=51.3gBe`

Regards.