The balanced equation :
2NaHCO₃⇒CO₂ + Na₂CO₃+H₂O
<h3>Further explanation</h3>
Given
Reaction
NaHCO(s) --> _CO2+_NaCO(s)+_H2O
Required
The balanced equation
Solution
Maybe the equation should be like this :
NaHCO₃⇒CO₂ + Na₂CO₃+H₂O
Give a coefficient
NaHCO₃⇒aCO₂ + bNa₂CO₃+cH₂O
Make an equation
Na, left=1, right=2b⇒2b=1⇒b=1/2
H, left=1, right=2c⇒2c=1⇒c=1/2
C, left=1, right=a+b⇒a+b=1⇒a+1/2=1⇒a=1/2
The equation becomes :
NaHCO₃⇒1/2CO₂ +1/2Na₂CO₃+1/2H₂O x2
2NaHCO₃⇒CO₂ + Na₂CO₃+H₂O
AgNO3(aq) + KCl (aq)-----> AgCl(s) + KNO3(aq)
Ag ^+(aq) + NO3^-(aq) + K^+ (aq) + cl^- (aq) ----> AgCl(s) + K^+(aq) + NO3^-(aq)
net ionic equation
Ag ^+(aq) + Cl^-(aq) ----> AgCl(s)
The percent yield : 73.5%
<h3>Further explanation</h3>
Given
Reaction
C+2H₂⇒CH₄
Required
The percent yield
Solution
mol of Carbon(as a limiting reactant) :
mol CH₄ based on C, and from equation mol ratio C : CH₄, so mol CH₄ = 8.3
Mass of Methane(theoretical yield) :
