How many gm of NACL are required to completely neutralize 100ml of 1Mof H2SO4?

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To determine the amount of NaCl (sodium chloride) required to completely neutralize 100 mL of 1 M (molar) H2SO4 (sulfuric acid), we need to consider the balanced chemical equation for the reaction between NaCl and H2SO4:

2 NaCl + H2SO4 → 2 HCl + Na2SO4

From the equation, we can see that 2 moles of NaCl react with 1 mole of H2SO4. Therefore, the molar ratio is 2:1.

Given that the concentration of H2SO4 is 1 M and the volume is 100 mL (which is equivalent to 0.1 L), we can calculate the moles of H2SO4 using the formula:

Moles = Molarity × Volume

Moles of H2SO4 = 1 M × 0.1 L = 0.1 moles

Now, using the molar ratio, we can find the moles of NaCl required:

Moles of NaCl = (2/1) × Moles of H2SO4 = 2 × 0.1 = 0.2 moles

The molecular weight (molar mass) of NaCl is approximately 58.44 g/mol. Therefore, the grams of NaCl required can be calculated as follows:

Grams of NaCl = Moles of NaCl × Molecular Weight of NaCl = 0.2 moles × 58.44 g/mol = 11.688 grams

So, approximately 11.688 grams of NaCl are required to completely neutralize 100 mL of 1 M H2SO4.

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