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How many grams of NaOH are required to neutralize 100 mL of a 6M HCl solution?

20 g
30 g
40 g
50 g

1 Answer

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Final answer:

To neutralize 100 mL of a 6M HCl solution, 0.6 moles of NaOH are required, which corresponds to 24 grams of NaOH, as calculated using the molar mass of NaOH. None of the multiple-choice options presented in the student's question matches this calculated value.

Step-by-step explanation:

To determine the amount of NaOH required to neutralize 100 mL of a 6M HCl solution, we can use the concept of molarity and the balanced chemical equation for the neutralization reaction between NaOH and HCl:

NaOH + HCl → NaCl + H2O

Since the reaction occurs in a 1:1 molar ratio, one mole of NaOH neutralizes one mole of HCl. Thus, we need to calculate the number of moles of HCl in 100 mL of a 6M solution:

Moles of HCl = Molarity × Volume (L) = 6 mol/L × 0.1 L = 0.6 moles

Therefore, we also need 0.6 moles of NaOH to fully neutralize the HCl. Now, we can calculate the mass of NaOH required using its molar mass (approximately 40.00 g/mol):

Mass of NaOH required = Moles × Molar mass = 0.6 moles × 40.00 g/mol = 24 g

Since 24 g is not one of the options provided in the question, ensure the calculation is correct, and if the test contains a typo or error, follow the instructions on handling such issues.

User Curtis Boylan
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