If 1.0 L of 1.0 M HCl spilled and needed to be neutralized, how many grams of the solid sodium carbonate (Na₂CO₃, 106 g/mol) would be required? A)53 g B)106 g C)1,060 g D)530 g E)212 g

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Quizplus · Accepted Answer

The neutralization reaction between HCl and Na2CO3 can be represented as: 2 HCl + Na2CO3 -> 2 NaCl + H2O + CO2. This shows that 2 moles of HCl react with 1 mole of Na2CO3. Since 1.0 L of 1.0 M HCl contains 1.0 mole of HCl, 0.5 moles of Na2CO3 are needed for neutralization. The molar mass of Na2CO3 is 106 g/mol, so 0.5 moles would weigh 53 g (0.5 * 106 g/mol = 53 g).

If 10 L of 1