Based solely on the amount of available carbon, how many grams of sodium oxalate Na2C2O4 can be obtained from 255 g of C6H6?
255g x (3*molar mass Na2C2O4/molar mass C6H6) = ?
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To determine the amount of sodium oxalate (Na2C2O4) that can be obtained from a given amount of benzene (C6H6) based on the available carbon, we need to use stoichiometry and the balanced chemical equation for the reaction.
The balanced chemical equation for the reaction between benzene and sodium oxalate can be written as follows:
C6H6 + 8NaOH + 2NaClO → 2Na2C2O4 + 6H2O + 2NaCl
From the equation, we can see that 1 mole of benzene reacts to form 2 moles of sodium oxalate.
First, we need to calculate the number of moles of benzene in 255 g of C6H6. To do this, we need to know the molar mass of benzene, which is 78.11 g/mol.
Number of moles of C6H6 = Mass of C6H6 / Molar mass of C6H6
= 255 g / 78.11 g/mol
≈ 3.27 mol
Next, we can use the stoichiometry from the balanced equation to determine the amount of sodium oxalate formed.
Number of moles of Na2C2O4 = 2 × Number of moles of C6H6
= 2 × 3.27 mol
≈ 6.54 mol
Finally, we can calculate the mass of sodium oxalate (Na2C2O4) obtained from the calculated moles.
Mass of Na2C2O4 = Number of moles of Na2C2O4 × Molar mass of Na2C2O4
= 6.54 mol × (2 × 22.99 g/mol + 2 × 12.01 g/mol + 4 × 16.00 g/mol)
≈ 6.54 mol × 133.99 g/mol
≈ 876.82 g
Therefore, based solely on the amount of available carbon in 255 g of benzene (C6H6), approximately 876.82 grams of sodium oxalate (Na2C2O4) can be obtained.