Answer:
a.36 g of water is produced.
b.64 g of [tex]O_{2}[/tex] is consumed.
Explanation:
The reaction is [tex]2H_{2} + O_{2}[/tex]⇒[tex]2H_{2}O[/tex]
a.
Given,
Weight of [tex]H_{2}[/tex] reacted = 4g
Weight of 1 mole of [tex]H_{2}[/tex] = 2[tex]\times[/tex]1 = 2g
Therefore no. of moles of [tex]H_{2}[/tex] reacted = [tex]\frac{4}{2}[/tex] = 2 moles;
Also given,
Weight of [tex]O_{2}[/tex] reacted = 32 g
Weight of 1 mole of [tex]O_{2}[/tex] = 2[tex]\times[/tex]16 = 32 g
Therefore no. of moles of [tex]O_{2}[/tex] reacted = [tex]\frac{32}{32}[/tex] = 1
We know that 2 moles of Hydrogen reacts with 1 mole of Oxygen to give 2 moles of water,
As we took 2 moles of Hydrogen and 1 mole of Oxygen,
Directly,from the equation we can tell 2moles of water will be produced.
Therefore no. of moles of [tex]H_{2} O[/tex] produced = 2
Weight of 1 mole of water = [tex]2\times 1+16[/tex] = 18
Therefore weight of [tex]H_{2}O[/tex] produced = [tex]2\times 18[/tex] = 36gm
b.
Given ,
72 g of [tex]H_{2}O[/tex] is produced.
So,
no. of moles of [tex]H_{2}O[/tex] produced =[tex]\frac{72}{18}[/tex] = 4 moles
From equation For every 2 moles of water formed , 1 mole of oxygen must be required.
So for producing 4 moles of water,
No. of moles of Oxygen required = 2 moles.
Therefore weight of [tex]O_{2}[/tex] reacted = [tex]2\times32[/tex] = 64 g
Method 2:
Given,
8 g of [tex]H_{2}[/tex] has reacted.
So,
no. of moles of [tex]H_{2}[/tex] reacted = [tex]\frac{8}{2}[/tex] = 4 moles.
From equation , we know that For every 2 moles of [tex]H_{2}[/tex] reacted,1 mole of [tex]O_{2}[/tex] will react.
Therefore,
No. of moles of [tex]O_{2}[/tex] that reacts with 4 moles of [tex]H_{2}[/tex] = [tex]2\times1[/tex] = 2 moles
Therefore the weight of [tex]O_{2}[/tex] reacted = [tex]2\times 32[/tex] = 64 g
