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
Answer:
120g
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
Sn + 2HF —> SnF2 + H2
Next, we shall determine the number of mole of HF needed to react with 3 moles of Sn.
From the balanced equation above, 1 mole of Sn reacted with 2 moles of HF.
Therefore, 3 moles of Sn will react with = 3 x 2 = 6 moles of HF.
Finally, we shall convert 6moles of HF to grams
This is illustrated below:
Number of mole of HF = 6moles
Molar Mass of HF = 1 + 19 = 20g/mol
Mass of HF =..?
Mass = number of mole x molar Mass
Mass of HF = 6 x 20
Mass of HF = 120g
Therefore, 120g of HF is needed to react with 3 moles of Sn
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