How many milliliters of 0.130 M H2SO4 are needed to neutralize 0.210 g of NaOH?
Express the volume in milliliters to three significant digits.
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First, we need to determine the number of moles of NaOH in 0.210 g.
The molar mass of NaOH is 22.99 g/mol + 16.00 g/mol + 1.01 g/mol = 39.00 g/mol.
Number of moles = mass / molar mass = 0.210 g / 39.00 g/mol = 0.00538 mol.
The balanced chemical equation for the neutralization reaction between H2SO4 and NaOH is:
H2SO4 + 2NaOH → Na2SO4 + 2H2O.
From the balanced equation, we can see that 1 mole of H2SO4 reacts with 2 moles of NaOH.
Therefore, the number of moles of H2SO4 needed to neutralize 0.00538 mol of NaOH is 0.00538 mol / 2 = 0.00269 mol.
The molarity of H2SO4 is 0.130 M, which means there are 0.130 moles of H2SO4 in 1 liter (1000 mL) of solution.
To find the volume of H2SO4 needed to neutralize 0.00269 mol, we can use the following equation:
Volume (mL) = moles of H2SO4 / molarity of H2SO4 = 0.00269 mol / 0.130 mol/L = 20.69 mL.
Therefore, 20.69 milliliters of 0.130 M H2SO4 are needed to neutralize 0.210 g of NaOH.
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