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Sodium carbonate and hydrochloric acid react according to the following equation:
Na2CO3+2 HCI-CO₂+ H2O +2 NaCl
How many liters of CO2 at STP can be produced from 13.5 g of sodium carbonate and an excess of HCI?
moles Na2CO3
moles of CO2
volume of CO2 |



Answer :

GinnyAnswer
To determine the volume of CO2 produced at STP from 13.5 g of sodium carbonate, we need to follow these steps: 1. Calculate the moles of sodium carbonate (Na2CO3): - The molar mass of Na2CO3 is: Na: 22.99 g/mol C: 12.01 g/mol O: 16.00 g/mol Molar mass of Na2CO3 = 2(22.99) + 12.01 + 3(16.00) = 105.99 g/mol - Moles of Na2CO3 = 13.5 g / 105.99 g/mol 2. Determine the moles of CO2 produced: - From the balanced chemical equation, 1 mole of Na2CO3 produces 1 mole of CO2. - Moles of CO2 = Moles of Na2CO3 3. Calculate the volume of CO2 at STP (Standard Temperature and Pressure): - 1 mole of any gas at STP occupies 22.4 liters. - Volume of CO2 = Moles of CO2 * 22.4 liters By following these steps, you can find the volume of CO2 produced at STP from the given mass of sodium carbonate. Remember to always check your calculations and units to ensure accuracy in your answer.

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