Question

When 1.229 g of sodium hydrogen carbonate (baking soda) is heated, it decomposes to 0.589 g of sodium carbonate. 2NaHCO3 (s) ? Na2CO3(s) + H2O(g) + CO2(g) Use the balanced chemical equation and other given information to determine the percent yield of sodium carbonate. Molar Mass of sodium hydrogen carbonate = 84.01 g/mol Molar Mass of sodium carbonate = 105.99 g/mol

          When 1.229 g of sodium hydrogen carbonate (baking soda) is heated, it decomposes to 0.589 g of sodium carbonate.

2NaHCO3 (s) ? Na2CO3(s) + H2O(g) + CO2(g)

Use the balanced chemical equation and other given information to determine the percent yield of sodium carbonate.

Molar Mass of sodium hydrogen carbonate = 84.01 g/mol

Molar Mass of sodium carbonate = 105.99 g/mol

When 1.229 g of sodium hydrogen carbonate (baking soda) is heated, it decomposes to 0.589 g of sodium carbonate.

2NaHCO3 (s) ? Na2CO3(s) + H2O(g) + CO2(g)

Use the balanced chemical equation and other given information to determine the percent yield of sodium carbonate.

Molar Mass of sodium hydrogen carbonate = 84.01 g/mol

Molar Mass of sodium carbonate = 105.99 g/mol

Added by Frank B.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

Instant Answer

Solved by Expert Adi S

09/01/2023

Step 1

First, we need to find the moles of sodium hydrogen carbonate (NaHCO3) in 1.229 g. moles of NaHCO3 = (mass of NaHCO3) / (molar mass of NaHCO3) moles of NaHCO3 = 1.229 g / 84.01 g/mol = 0.01463 mol Show more…

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When 1.229 g of sodium hydrogen carbonate (baking soda) is heated, it decomposes to 0.589 g of sodium carbonate 2NaHCO3 (s) → Na2CO3(s) + H2O(g) + CO2(g). Use the balanced chemical equation and other given information to determine the percent yield of sodium carbonate. Molar Mass of sodium hydrogen carbonate: 84.01 g/mol. Molar Mass of sodium carbonate: 105.99 g/mol.