Answer:
74.4 ml
Explanation:
C₆H₈O₇(aq) + 3NaHCO₃(s) => Na₃C₆H₅O₃(aq + 3CO₂(g) + 3H₂O(l)
Given 15g = 15g/84g/mol = 0.1786mole Sodium Bicarbonate
From equation stoichiometry 3moles NaHCO₃ is needed for each mole citric acid or, moles of citric acid needed is 1/3 of moles sodium bicarbonate used.
Therefore, for complete reaction of 0.1786 mole NaHCO₃ one would need 1/3 of 0.1786 mole citric acid or 0.0595 mole H-citrate.
The question is now what volume of 0.8M H-citrate solution would contain 0.0595mole of the H-citrate? This can be determined from the equation defining molarity. That is => Molarity = moles solute / Liters of solution
=> Volume (Liters) = moles citric acid / Molarity of citric acid solution
=> Volume needed in liters = 0.0.0595 mole/0.80M = 0.0744 Liters or 74.4 ml
Answer:
The answer to your question is 432 g of CO₂
Explanation:
Data
CaCO₃ = 983 g
CaO = 551 g
CO₂ = ?
Balanced reaction
CaCO₃ (s) ⇒ CaO (s) + CO₂ (g)
This reaction is balanced, to solve this problem just remember the Lavoisier Law of conservation of mass that states that the mass of the reactants is equal to the mass of the products.
Mass of reactants = Mass of products
Mass of CaCO₃ = Mass of CaO + Mass of CO₂
Solve for CO₂
Mass of CO₂ = Mass of CaCO₃ - Mass of CaO
Mass of CO₂ = 983 g - 551 g
Simplification
Mass of CO₂ = 432 g
Put the salt and sand in some water.
the salt will not be visible but the sand will
now strain the the sand from the water
now boil the water and now the water will boil away and now you will just have salt left.
