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

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.