A.) How many moles of NaOH will be needed to neutralize 0.49 mol of H2SO4? The f

Question

A.) How many moles of NaOH will be needed to neutralize 0.49 mol of H2SO4? The following equation is NON-balanced: NaOH + H2SO4 --> Na2SO4 + H2O

B.)A sample contains both NaOH and NaCl. 0.500 g of this sample was dissolved in water to make a 20.0 mL solution and then this solution was titrated by 0.500 mol/L HCl solution. If 20.8 mL of HCl was used to reach the end point, what is the mass % of NaOH in the sample?

NaOH + HCl --> NaCl + H2O

Note:

1. Keep 3 sig figs for your final answer.

2. Answer as percentage.

Explanation / Answer

A) 2NaOH + H2SO4 -------------> Na2SO4 + 2H2O

according to balanced reaction

1 mole H2SO4 reacts with 2 moles of NaOH

0.49 moles of H2SO4 reacts with 0.49 x 2 = 0.98 moles of NaOH required.

moles of NaOH required = 0.98 moles

B) moles of HCl consumed = 0.5 x 20.8 / 1000 = 0.0104 moles

so 0.0104 moles of NaOH must be present

moles = W/MW

W = moles x MW = 0.0104 x 40 = 0.416 g

mass of NaOH = 0.416 g

mass % = (0.416 / 0.5) x 100

mass % = 83.2

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