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A 1.0 kg bottle of sodium carbonate (Na2CO3, 106.0 g/mol) is available to clean up 5.00 liters of spilled concentrated aqueous hydrochloric acid (9.75 M). Is this enough sodium carbonate to neutralize the acid according to the following reaction?
2 HCl (aq) + Na2CO3 (s)  2NaCl (aq) + CO2 (g) + H2O (l)
(1) No, there is approximately 40% too small amount of sodium carbonate needed.
(2) Yes, there is approximately 80% more than what is needed.
(3) No, there is approximately 60% too small amount of sodium carbonate needed.
(4) Yes, there is exactly enough sodium carbonate, but no excess.
(5) No, there is approximately 20% too small amount of sodium carbonate needed.

Respuesta :

Answer:

The correct answer is option 4, that is, there is exactly enough sodium carbonate.

Explanation:

Based on the given question, the reaction will be,

2 HCl (aq) + Na2CO3 (s) ⇒ 2 NaCl (aq) + CO2 (g) + H2O (l)

Therefore, for neutralizing 2 moles of HCl, one mole of Na2CO3 is required.

No of moles present in 1 Kg or 1000 grams of Na2CO3 will be,

Moles = Weight/Molecular mass of Na2CO3

Moles = 1000 / 106 = 9.43

Thus, 9.43 moles of Na2CO3 is present.

No of moles present in 1 liter of 9.75 M HCl is 9.75.

No. of moles present in 5 Liters of HCl (9.75 M),

= 5 × 9.75 = 48.75

Thus, for 2 moles of HCl 1 mole of Na2CO3 is required. Now for 48.75 moles of HCl, the moles required of Na2CO3 is 9.75. Therefore, for complete neutralization, the moles of Na2CO3 required is 9.75, and the present moles is 9.43.

Hence, there is exactly enough sodium carbonate.

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