Exactly 0.1120 g of pure Na2CO3 was dissolved in 100.0 mL of 0.0497 M HClO4. a)

Question

Exactly 0.1120 g of pure Na2CO3 was dissolved in 100.0 mL of 0.0497 M HClO4. a) What mass in grams of CO2 was evolved? The balanced equation I set up is: Na2CO3 + 2HClO4 --> NaClO4 + CO2 + H2O Now what? b) What was the molarity of the excess reactant (HCl or Na2CO3)?

Explanation / Answer

Na2CO3 + 2HClO4 -> 2NaClO4 + CO2 +H2O a)No. of moles of Na2CO3 present is 0.112/106 and required amount of HClO4 is 2*0.112/106 = 0.00211 moles where given no of moles of HClO4 is 0.0049 moles so HClO4 is excess continue the equation with Na2CO3 1 mole of Na2CO3 gives one mole of CO2 o.112g of Na2CO3 gives 0.112*44/106 = 0.044g of CO2 b)Now the excess of HClO4 is (0.0049-0.00211) moles = 0.00279 moles Molarity = 0.00279/0.1 = 0.0279 M

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