when excess carbon dioxide is passed into sodium hydroxide solution, sodium carbonate solution is formed. this can be crystallised out as Na2CO3.10H2O . what mass of crystals would be produced from 5g of sodium hydroxide in excess water
To determine the mass of sodium carbonate crystals produced, we need to use stoichiometry and mole concept.
1. Write the balanced equation for the reaction:
CO2 + 2NaOH + H2O → Na2CO3·10H2O
2. Convert the given mass of sodium hydroxide (NaOH) to moles:
Molar mass of NaOH = 22.99 + 16.00 + 1.008 = 39.998 g/mol
Moles of NaOH = 5 g / 39.998 g/mol ≈ 0.125 mol
3. Determine the mole ratio between NaOH and Na2CO3·10H2O from the balanced equation:
From the equation, we see that 1 mole of NaOH reacts with 1 mole of Na2CO3·10H2O.
4. Calculate the moles of Na2CO3·10H2O that can be formed:
Moles of Na2CO3·10H2O = 0.125 mol
5. Convert moles of Na2CO3·10H2O to grams:
Molar mass of Na2CO3·10H2O = 22.99*2 + 12.01 + 16.00*3 + 10*(1.008*2 + 16.00) = 286.14 g/mol
Mass of Na2CO3·10H2O = Moles of Na2CO3·10H2O * Molar mass of Na2CO3·10H2O
= 0.125 mol * 286.14 g/mol
≈ 35.7685 g
Therefore, approximately 35.7685 grams of sodium carbonate crystals (Na2CO3·10H2O) would be produced from 5 grams of sodium hydroxide (NaOH) in excess water.