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Directions: Using the definition of molarity, the given balanced equations, and stoichiometry, solve the following problems.

Sodium chloride solution and water react to produce sodium hydroxide and chlorine gas according to the following balanced equation: 2NaCl(aq) + 2H20(l) <-> 2Na0H(aq) + Cl2(g)
a. How many liters of 0.4 M sodium chloride do you need in order to have 3.0 moles of chlorine gas?
b. Find the number of moles of water needed to produce 3.0 L of chlorine gas at STP.

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Answer:

15L of 0.40M NaCl(aq) solution

Explanation:

2NaCl(aq) + 2H₂O(l) → 2NaOH(aq) + Cl₂(g)

2Na⁺(aq) + 2Clˉ(aq) + 2H₂O(l) → 2Na⁺(aq) + 2OHˉ(aq) + Cl₂(g)

Na⁺(aq) is a spectator ion in the given reaction and does not enter into the reaction process…

Net Ionic Equation is then 2Clˉ(aq) + 2H₂O(l) → 2OHˉ(aq) + Cl₂(g)

From Rxn, 2 moles Clˉ(aq) is needed to produce 1 mole of  Cl₂(g)

Therefore, 6 moles Clˉ(aq) is needed to produce 3 moles of Cl₂(g)

That is, 6 moles NaCl(aq) → 6 moles Clˉ(aq) = 0.40M x V(NaCl)liters  

V(NaCl) liters = 6 moles Clˉ(aq)/(0.40mole/liter) = 15 liters of 0.40M NaCl(aq)

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