Explanation:
The substance that is in excess that doesn't get used up as a reactant is called limiting reactant.
Mass to moles
5.2 mol/Ca(no3)2 to mol
5.2 mol/Ca(no3)2 / molar mass
5.2 mol/Ca(no3)2 / 164.1= 0.032 g/Ca(no3)2
Answer:
The answer to your question is 8.21 g of H₂O
Explanation:
Data
mas of water = ?
mass of hydrogen = 4.6 g
mass of oxygen = 7.3 g
Balanced chemical reaction
2H₂ + O₂ ⇒ 2H₂O
Process
1.- Calculate the atomic mass of the reactants
Hydrogen = 4 x 1 = 4 g
Oxygen = 16 x 2 = 32 g
2.- Calculate the limiting reactant
Theoretical yield = H₂/O₂ = 4 / 32 = 0.125
Experimental yield = H₂/ O₂ = 4.6/7.3 = 0.630
From the results, we conclude that the limiting reactant is Oxygen because the experimental yield was higher than the theoretical yield.
3.- Calculate the mass of water
32 g of O₂ ---------------- 36 g of water
7.3 g of O₂ --------------- x
x = (7.3 x 36) / 32
x = 262.8 / 32
x = 8.21 g of H₂O
