# When sodium carbonate reacts with hydrochloric acid, the carbonic acid that is formed immediately breaks down into carbon dioxide and water. What mass of sodium carbonate would have been originally...

When sodium carbonate reacts with hydrochloric acid, the carbonic acid that is formed immediately breaks down into carbon dioxide and water. What mass of sodium carbonate would have been originally present if 5.0 L of carbon dioxide was produced? NA2CO3--> 2NaCL + CO2+H2O

llltkl | Student

When sodium carbonate reacts with hydrochloric acid, the carbonic acid that is formed immediately breaks down into carbon dioxide and water. What mass of sodium carbonate would have been originally present if 5.0 L of carbon dioxide was produced?

The balanced chemical equation for the reaction of sodium carbonate  and hydrochloric acid is:

`Na_2CO_3 +2HCl = 2NaCl + CO_2 (g)+H_2O`

From the stoichiometry of this reaction, it follows that:

At NTP, 1 gram-mole, i.e. 22.4 litre `CO_2 (g)` is produced from (2*23+12+3*16), i.e. 106 g `Na_2CO_3` .

Therefore 5.0 L `CO_2` (g) will be produced from `(106*5.0)/22.4=23.66` g `Na_2CO_3`

ayl0124 | Student

Use stoichiometry to solve this problem. The units cross each other out, leaving you with "g Na2CO3."

`5.0"L(CO_2)" = (1"mol(CO_2)")/(22.4"L(CO_2)") = (1"mol(Na_2CO_3)")/(1"mol(CO_2)") = (105.99"g(Na_2CO_3)")/(1"mol(Na_2CO_3)") = 23.66"g(Na_2CO_3)" `

You would need 24 grams of Na2CO3 to produce 5.0 L of CO2. Note that you need two sig figs in your answer because you began with two sig figs.