Ca(OH)2+Na2CO3> 2NaOH + CaCO3
50g hydroxide and 50g sodium carbonate are reacted. How many grams sodium hydroxide will be made?
To determine how many grams of sodium hydroxide (NaOH) will be made from the reaction between 50g of calcium hydroxide (Ca(OH)2) and 50g of sodium carbonate (Na2CO3), we need to first find the limiting reactant.
1. Calculate the molar mass of each compound:
- Ca(OH)2: 40.08g/mol (Ca) + 2(1.01g/mol (H) + 16.00g/mol (O)) = 74.10g/mol
- Na2CO3: 2(22.99g/mol (Na) + 12.01g/mol (C) + 3(16.00g/mol (O))) = 105.99g/mol
2. Determine the number of moles for each reactant:
- Moles of Ca(OH)2 = 50g / 74.10g/mol = 0.675 mol
- Moles of Na2CO3 = 50g / 105.99g/mol = 0.472 mol
3. Write the balanced equation for the reaction:
Ca(OH)2 + Na2CO3 → 2NaOH + CaCO3
4. Use the stoichiometric coefficients from the balanced equation to determine the limiting reactant. The reactant that produces fewer moles of the desired product, in this case NaOH, is the limiting reactant.
- According to the balanced equation, 1 mole of Ca(OH)2 reacts to produce 2 moles of NaOH.
- According to the balanced equation, 1 mole of Na2CO3 reacts to produce 2 moles of NaOH.
Since both reactants have a stoichiometric ratio of 1:2 with NaOH, we can see that the limiting reactant is Na2CO3 because it produces fewer moles of NaOH. We have 0.472 mol of Na2CO3, which will determine the maximum amount of NaOH that can be produced.
5. Calculate the number of moles of NaOH that can be produced:
- Moles of NaOH = 2 × moles of Na2CO3 = 2 × 0.472 mol = 0.944 mol
6. Finally, calculate the mass of NaOH produced using the molar mass of NaOH:
- Mass of NaOH = Moles of NaOH × Molar mass of NaOH
- Mass of NaOH = 0.944 mol × 39.997 g/mol = 37.8 g
Therefore, approximately 37.8 grams of sodium hydroxide (NaOH) will be produced when 50 grams of calcium hydroxide (Ca(OH)2) and 50 grams of sodium carbonate (Na2CO3) are reacted.
To solve this problem, you'll need to use stoichiometry, which is a method of using balanced chemical equations to calculate the quantities of reactants and products in a chemical reaction.
First, let's determine the molar masses of the compounds involved:
- The molar mass of Ca(OH)2 (calcium hydroxide) can be calculated:
- Ca: 40.08 g/mol
- O: 16.00 g/mol (x2 because there are two oxygen atoms)
- H: 1.01 g/mol (x2 because there are two hydrogen atoms)
- Total molar mass: 40.08 + 16.00(2) + 1.01(2) = 74.09 g/mol
- The molar mass of Na2CO3 (sodium carbonate) can be calculated:
- Na: 22.99 g/mol (x2 because there are two sodium atoms)
- C: 12.01 g/mol
- O: 16.00 g/mol (x3 because there are three oxygen atoms)
- Total molar mass: 22.99(2) + 12.01 + 16.00(3) = 105.99 g/mol
Based on the balanced chemical equation, we can see that 1 mole of Ca(OH)2 reacts with 1 mole of Na2CO3 to produce 2 moles of NaOH and 1 mole of CaCO3.
Now, let's calculate the moles of Ca(OH)2 and Na2CO3:
- Moles of Ca(OH)2 = (mass of Ca(OH)2 / molar mass of Ca(OH)2) = 50g / 74.09 g/mol
- Moles of Na2CO3 = (mass of Na2CO3 / molar mass of Na2CO3) = 50g / 105.99 g/mol
Since the reaction ratio is 1:1 for Ca(OH)2 and Na2CO3, the moles of Ca(OH)2 and Na2CO3 will be the same.
Finally, we can calculate the moles (and mass) of NaOH produced:
- Moles of NaOH = 2 × Moles of Ca(OH)2 = 2 × (50g / 74.09 g/mol)
- Mass of NaOH = Moles of NaOH × molar mass of NaOH
Perform the calculations to find the answer.