Answer:
mass of NaHCO₃ = 0.126 g
Step-by-step explanation:
balanced chemical equation
H₃C₆H₅O₇(aq) + 3 NaHCO₃(aq) → 3 H₂O(l) + 3 CO₂(g) + Na₃C₆H₅O₇(aq).
given data
moles of citric acid = 1 mol
mas of citric acid = 100 mg
mas of citric acid = 0.1 g
moles of sodium bicarbonate = 3 mol
mass of sodium bicarbonate = ?
Solution
1st we will find out the mole ratio of H₃C₆H₅O₇ and NaHCO₃ from balanced chemical equation
H₃C₆H₅O₇ : NaHCO₃
1 : 3
Now we find out number of moles of citric acid
moles = mass / molar mass
moles = 0.1 g / 192.124 g/mol
moles = 0.0005 mol
now we find out moles (x) of sodium bicarbonate needed for 0.0005 mol of citric acid
from the balanced chemical equation the mole ratios are:
1 : 3
0.0005 mol : x
Cross multiply these ratios
3 × 0.0005 = 1x
0.0015 = 1 x
x = 0.0015 mol
Now we will find out the mass of sodium bicarbonate
mass = moles × molar mass
mass of NaHCO₃ = 0.0015 mol × 84.007 g/mol
mass of NaHCO₃ = 0.126 g