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

25.0 mL of 0.50 M NaOH were combined with 14.7 mL of 0.35 M H 2 SO 4 . Determine the moles of the excess reactant remaining and its new concentration. 2NaOH + H 2 SO 4  2H 2 O + Na 2 SO 4. 25.0 mL. 14.7 mL. Just pick a product.

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

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  1. 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

  2. 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

  3. 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 0.0103 = mol H2O Theoretical yield? Limiting Reactant?

  4. 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 0.0103 = mol H2O 0.0125 – 0.0103 = 0.0022 moles of H2O that can be formed from the excess NaOH.

  5. 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

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