If 930 mL of 0.0185 M aqueous Na 2 CO 3 and800 mL of 0.0413 M aqueous HCl are re

Question

If 930 mL of 0.0185 M aqueous Na2CO3 and800 mL of 0.0413 M aqueous HCl are reacted, what volume (L) ofgaseous CO2 measured at 0.5 atm pressure and 309 Kis produced.

Na2CO3(aq) + 2 HCl(aq) CO2(g) + 2 NaCl(aq) + H2O(l)
Does this problem pertain to thelaw of partial pressures? If so, do I just find the moles of Na2CO3and HCI, add them together and then solve forV=nRT/P?
for finding moles of Na2CO3 andHCI, do I convert mL to L? or leave it as is?
Does this problem pertain to thelaw of partial pressures? If so, do I just find the moles of Na2CO3and HCI, add them together and then solve forV=nRT/P?
for finding moles of Na2CO3 andHCI, do I convert mL to L? or leave it as is?

Explanation / Answer

You have to find the moles of sodium carbonate andhydrochloric acid by converting ml to L. Moles = molarity * volume In the balanced equation, 2 moles of HCl react with 1 mole ofsodium carbonate. From the moles, we find out which is the limiting reactantbased on the stoichiometric ratio above. Based on the ratio of moles of reactants, the moles of carbondioxide being produced can be obtained. Then substitute in V = nRT / P This gives the final volume of gaseous CO2 produced.

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