25.0 mL of 0.50 M NaOH were combined with 14.7 mL of 0.35 M

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25.0 mL of 0.50 M NaOH were combined with 14.7 mL of 0.35 M H2SO4. Determine the moles of the excess reactant remaining and its new concentration. 2NaOH + H2SO4  2H2O + Na2SO4 25.0 mL 14.7 mL Just pick a product

25.0 mL of 0.50 M NaOH were combined with 14.7 mL of 0.35 M H2SO4. Determine the moles of the excess reactant remaining and its new concentration. 2NaOH + H2SO4  2H2O + Na2SO4 25.0 mL 14.7 mL Let’s choose H2O

25.0 mL of 0.50 M NaOH were combined with 14.7 mL of 0.35 M H2SO4. Determine the moles of the excess reactant remaining and its new concentration. 2NaOH + H2SO4  2H2O + Na2SO4 25.0 mL 14.7 mL ? moles NaOH 25.0 mL Based on: NaOH = mol H2O 0.0125 Based on: H2SO4 = mol H2O 0.0103 Theoretical yield? Limiting Reactant?

25.0 mL of 0.50 M NaOH were combined with 14.7 mL of 0.35 M H2SO4. Determine the moles of the excess reactant remaining and its new concentration. 2NaOH + H2SO4  2H2O + Na2SO4 25.0 mL 14.7 mL ? moles NaOH 25.0 mL Based on: NaOH = mol H2O 0.0125 Based on: H2SO4 = mol H2O 0.0103 0.0125 – 0.0103 = 0.0022 moles of H2O that can be formed from the excess NaOH.

SO: 0.0022 mol H2O = mole NaOH 0.0022 NOW: 0.0022 mol NaOH 0.055 25.0 mL of 0.50 M NaOH were combined with 14.7 mL of 0.35 M H2SO4. Determine the moles of the excess reactant remaining and its new concentration. 2NaOH + H2SO4  2H2O + Na2SO4 25.0 mL 14.7 mL ? moles 0.0125 – 0.0103 = 0.0022 moles of H2O that can be formed in excess from the excess NaOH. SO: 0.0022 mol H2O = mole NaOH 0.0022 NOW: 0.0022 mol NaOH 0.055 39.7 x 10-3 Lsolution