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.
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
1 answer
1 answer