Answer: 15.1 grams
Given reaction:
Na2CO3 + Ca(OH)2 → 2NaOH + CaCO3
Mass of Na2CO3 = 20.0 g
Molar mass of Na2CO3 = 105.985 g/mol
# moles of Na2CO3 = 20/105.985 = 0.1887 moles
Based on the reaction stoichiometry: 1 mole of Na2CO3 produces 2 moles of NaOH
# moles of NaOH produced = 0.1887*2 = 0.3774 moles
Molar mass of NaOH = 22.989 + 15.999 + 1.008 = 39.996 g/mol
Mass of NaOH produced = 0.3774*39.996 = 15.09 grams
Explanation:
Answer:
Both i belive Health class
Answer:
The mass of 
in the container is 2.074 gram
Explanation:
Given:
Volume of 
lit
Equilibrium constant 
The reaction in which is produced
⇄ 
Here equal moles of and 
is formed.
From the formula of equilibrium constant,
M
Above value shows,
So in 2 L no. moles of = 
moles.
So mass of 0.122 mole of is = 
g
Therefore, the mass of in the container is 2.074 gram
