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ICE Challenge.

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Presentation on theme: "ICE Challenge."— Presentation transcript:

1 ICE Challenge

2 1. 25. 0 mL of 0. 100 M NaOH, 10. 0 mL 0. 200 M KOH, and 20. 0 mL of 0
mL of M NaOH, 10.0 mL M KOH, and mL of M H2SO4 are poured into the same beaker. What is the resulting concentration of the excess acid or base? L x mole NaOH = mol L L x mole KOH = mol = mol Total Base L x mole H2SO4 = mol Total Acid 2XOH + H2SO4 → X2SO4 + 2HOH I mol mol C mol mol E mol Total Volume = mL mL mL = mL Molarity Bases = mol = M L

3 mL of M HCl and mL M HNO3, react with excess CaCO3. What is the resulting volume in mLof CO2 produced at STP. 2HCl CaCO3 → CO2 + CaCl2 H2O 0.250 L HCl x mole x 1 mole CO2 x L x mL = mL 1 L 2 mole HCl 1 mole 1L 2HNO CaCO3 → CO2 + Ca(NO3)2 + H2O 0.100 L HNO3 x mole x 1 mole CO2 x L x mL = mL 1 L 2 mole HNO3 1 mole 1L Total CO2 = mL


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