Respuesta :
Hey there!:
Molar mass:
HNO3 = 63.01 g/mol
H2O = 18.01 g/mol
By the stoichiometry of the reaction:
Ca(OH)2 + 2HNO3 -> Ca(NO3)2+ 2H2O
( 2 * 63.01 ) g HNO3 --------------- ( 2 * 18.01 ) g H2O
mass of HNO3 ---------------------- 22.8 g H2O
mass of HNO3 = ( 22.8 * 2 * 63.01 ) / (2 * 18.01 )
mass of HNO3 = 2873.256 / 36.02
mass of HNO3 = 79.768 g
Therefore:
1 mole HNO3 ------------------ 63.01 g
moles HNO3 ------------------ 79.768 g
moles HNO3 = ( 79.768 * 1 ) / ( 63.01 )
moles HNO3 = 79.768 / 63.01
=> 1.265 moles of HNO3
Hope that helps!
Answer:
1.27 mol HNO₃
Explanation:
We know we will need a balanced chemical equation with masses, moles, and molar masses.
1. Gather all the information in one place:
M_r: 18.02
Ca(OH)₂ + 2HNO₃ ⟶ Ca(NO₃)₂ + 2H₂O
m/g: 22.8
2. Calculate the moles of H₂O
Moles of H₂O = 22.8 g H₂O × (1 mol H₂O/18.02 g H₂O)
= 1.265 mol H₂O
3. Calculate the moles of HNO₃
The molar ratio is 2 mol HNO₃/2 mol H₂O.
Moles of HNO₃ = 1.265 mol H₂O × (2 mol HNO₃/2 mol H₂O)
= 1.27 mol HNO₃