Answer:
18 g is the mass produced by 4 g of H₂ and 16 g of O₂
Explanation:
The reaction is:
2H₂ + O₂ → 2H₂O
So, let's find out the limiting reactant as we have both data from the reactants.
Mass / Molar mass = moles
4 g/ 2g/m = 2 moles H₂
16g / 32 g/m = 0.5 moles O₂
2 moles of hydrogen react with 1 mol of oxygen, but I have 0.5, so the O₂ is the limiting.
1 mol of O₂ produces 2 mol of water.
0.5 mol of O₂ produce (0.5 .2)/1 = 1 mol of water.
1 mol of water weighs 18 grams.
Answer:
18 grams of [tex]H_2O[/tex]
Explanation:
The balanced equation of the reaction is:
[tex]H_2+\frac{1}{2}O_2 -->H_2O[/tex]
From the balanced equation we can say 1 mole of H2 reacts with 0.5 moles of O2 to give one mole of H2O.
Number of moles of H2 = [tex]\frac{Given\ mass}{Molar\ mass}=\frac{4}{2}=2\ moles[/tex]
Number of moles of O2 = [tex]\frac{Given\ mass}{Molar\ mass}=\frac{16}{32}=0.5\ moles[/tex]
We have 2 moles H2 and 0.5 moles of O2.
Not all H2 reacts because the amount of O2 is limited.
Since only 0.5 moles of O2 is available only 1 mole of H2 reacts according to the balanced equation.
Hence 1 mole of H2O is formed which is 18 grams.
