So im in 10th grade and I got a chemistry lab that i dont get at all
Steps-
1) Balancing Equation when 22mL of Hydrochloric Acid (0.5M) is reacted to 1.5g of Sodium Bicarbonate,
2)Amount of CO2 Produced from this reaction.
3)The limiting reagent in this experiment.
I would really appreciate if you show me all the steps and calculations
Concentrations-
22mL of Hydrochloric Acid (0.5M)
1.5g Sodium Bicarbonate
PLS PLS PLS PLS PLS PLS PLS HELP - URGENT CHEM LAB ;/ - Moles and balancing equations?
2016-06-09 5:44 pm
回答 (1)
2016-06-09 6:03 pm
1)
HCl(aq) + NaHCO₃(s) → NaCl(aq) + H₂O(l) + CO₂(g)
2)
From Wikipedia: Molar mass of NaHCO₃ = 84.0066 g/mol
When HCl completely reacts :
(0.5 mol HCl/L) × (22/1000 L) × (1 mol CO₂ / 1 mol HCl) = 0.011 mol CO₂
When NaHCO₃ completely reacts :
[(1.5 g NaHCO₃) / (84.0066 g NaHCO₃ / 1 mol NaHCO₃)] × (1 mol CO₂ / 1 mol NaHCO₃) = 0.0179 mol CO₂
0.011 mol CO₂ < 0.0179 mol CO₂
Amount of CO₂ produced = 0.011 mol
3.
When completely reacts, HCl produces a smaller amount of CO₂.
Hence, HCl is the limiting reagent.
HCl(aq) + NaHCO₃(s) → NaCl(aq) + H₂O(l) + CO₂(g)
2)
From Wikipedia: Molar mass of NaHCO₃ = 84.0066 g/mol
When HCl completely reacts :
(0.5 mol HCl/L) × (22/1000 L) × (1 mol CO₂ / 1 mol HCl) = 0.011 mol CO₂
When NaHCO₃ completely reacts :
[(1.5 g NaHCO₃) / (84.0066 g NaHCO₃ / 1 mol NaHCO₃)] × (1 mol CO₂ / 1 mol NaHCO₃) = 0.0179 mol CO₂
0.011 mol CO₂ < 0.0179 mol CO₂
Amount of CO₂ produced = 0.011 mol
3.
When completely reacts, HCl produces a smaller amount of CO₂.
Hence, HCl is the limiting reagent.
收錄日期: 2021-04-18 14:57:42
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