Answer: [tex]O_2[/tex] is the limiting reagent, 180 g of water is produced
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]\text{Moles of} H_2=\frac{24.0g}{2g/mol}=12moles[/tex]
[tex]\text{Moles of} O_2=\frac{160.0g}{32g/mol}=5moles[/tex]
[tex]2H_2+O_2\rightarrow 2H_2O(g)[/tex]
According to stoichiometry :
1 mole of [tex]O_2[/tex] require 2 moles of [tex]H_2[/tex]
Thus 5 moles of [tex]O_2[/tex] will require=[tex]\frac{2}{1}\times 5=10moles[/tex] of [tex]H_2[/tex]
Thus [tex]O_2[/tex] is the limiting reagent as it limits the formation of product and [tex]H_2[/tex] is the excess reagent.
As 1 mole of [tex]O_2[/tex] give = 2 moles of [tex]H_2O[/tex]
Thus 5 moles of [tex]O_2[/tex] give =[tex]\frac{2}{1}\times 5=10moles[/tex] of [tex]H_2O[/tex]
Mass of [tex]H_2O=moles\times {\text {Molar mass}}=10moles\times 18g/mol=180g[/tex]
Thus 180 g of [tex]H_2O[/tex] will be produced from the given masses of both reactants.
