Question

determine how much zinc of 50% purity will be required to prepare 5 gram of gaseous hydrogen from dilute sulphuric acid (Zn=65.5,O=16,S=32,H=1)​

Answer 1

Given: 50% pure zinc is given and $5gm$ of gaseous hydrogen is to be prepared from dilute sulphuric acid.

To find: The amount of $50%$% pure zinc needed to make $5gm$ of gaseous hydrogen

Solution:

The chemical equation for the process of making hydrogen from zinc and sulphuric acid is:

$Zn+H_2SO_4$ ↔  $ZnSO_4+H_2$

The molar mass of $Zn=65.5gm$

The molar mass of $H_2=2gm$

As per the equation,

$65.5gm$ of Zinc is required to produce $2g$ of hydrogen gas

The purity of the zinc sample is $50%$% that is the actual gain of the product will be $50$% of the theoretical gain,

∴ The amount of hydrogen produced by $65.5gm$ of zinc $=(2*50)/100=1gm$

Zinc sample needed to yield $1gm$ of hydrogen $=65.5gm$

∴ Amount of zinc needed to yield $5gm$ of hydrogen $=65.5*5=327.5gm$

Final answer:

The amount zinc needed to yield 5 grams of gaseous hydrogen is $327.5 gm$