Answer:
81.7 %
Explanation:
Equation of the decomposition of NaHCO₃
2 NaHCO₃ → Na₂CO₃ + CO₂ + H₂O
2 mole of NaHCO₃ yielded 1 mole of CO₂ and 1 mole of H₂O
molar mass of CO₂ = 44 g
molar mass of H₂O = 18 g
1 mole of CO₂ : 1 mole H₂O = ( mass of CO₂ / molar mass of CO₂) : ( mass of H₂O / molar mass of H₂O
also mass of CO₂ + mass of H₂O = ( 0.654 - 0.456 ) g = 0.198 g
1 = ( mass of CO₂/ 44g) : ( mass of H₂O / 18g)
44 mass of H₂O = 18 mass of CO₂
1 mass of CO₂ = 44 / 18 mass of H₂O
substitute into equation 1
mass of CO₂ + mass of H₂O = 0.198 g
2.44 mass of H₂O + mass of H₂O = 0.198 g
3.44 mass of H₂O = 0.198 g
mass of H₂O = 0.198 g / 3.44 = 0.0576 g
mass of CO₂ = 0.198 g - 0.0576 g = 0.140 g
2 mole of NaHCO₃ yielded 1 mole of CO₂
168 g of NaHCO₃ yielded 44 g of CO₂
unknown mass of NaHCO₃ yielded 0.140 g CO₂
unknown mass of NaHCO₃ = 168 g × 0.140 g / 44 g = 0.535 g
mass percent of NaHCO₃ = 0.535 g / 0.654 g = 81.7 %
Answer:
Explanation:
Mass of impure NaHCO3 = 0.654g
Mass of residue= 0.456g
Mass loss of NaHCO3= 0.654-0.456= 0.198g
Balanced reaction equation:
2NaHCO3(s)-------> Na2CO3(s) + H2O(g)+CO2(g)
Note CO2 and water vapour produced in the decomposition combines to give H2CO3
84g of NaHCO3 yields 62g of H2CO3
Xg of NaHCO3 yields 0.198g of H2CO3
Therefore X= 84 × 0.198/ 62
=0.268g
Mass%= 0.268/0.654 × 100
=41% NaHCO3
