how many grams of sodium oxalate, Na2C2O4, could be obtained from 213 g of C6H6?
zero. Do you have some particular reaction in mind?
bhujgy
To determine the amount of sodium oxalate that could be obtained from a given amount of benzene (C6H6), we need to use stoichiometry, which is the relationship between the amounts of reactants and products in a chemical reaction.
1. Write the balanced equation for the reaction between benzene (C6H6) and sodium hydroxide (NaOH):
C6H6 + 2NaOH -> Na2C2O4 + 4H2O
2. Calculate the molar mass of benzene (C6H6) and sodium oxalate (Na2C2O4):
Molar mass of C6H6 = (12.01 g/mol * 6) + (1.01 g/mol * 6) = 78.11 g/mol
Molar mass of Na2C2O4 = (22.99 g/mol *2) + (12.01 g/mol * 2) + (16.00 g/mol * 4) = 134.01 g/mol
3. Convert the given mass of benzene (C6H6) to moles:
Moles of C6H6 = Mass of C6H6 / Molar mass of C6H6
= 213 g / 78.11 g/mol
= 2.724 mol
4. Use the stoichiometric coefficients from the balanced equation to determine the mole-to-mole ratio between C6H6 and Na2C2O4:
1 mole of C6H6 reacts to produce 1 mole of Na2C2O4
5. Calculate the moles of Na2C2O4 that could be formed:
Moles of Na2C2O4 = Moles of C6H6 * (1 mole of Na2C2O4 / 1 mole of C6H6)
= 2.724 mol * (1/1)
= 2.724 mol
6. Convert the moles of Na2C2O4 to grams:
Mass of Na2C2O4 = Moles of Na2C2O4 * Molar mass of Na2C2O4
= 2.724 mol * 134.01 g/mol
= 365.49 g
Therefore, approximately 365.49 grams of sodium oxalate (Na2C2O4) could be obtained from 213 grams of benzene (C6H6).