Question

The following reaction is stoichiometric as written C4H9Cl + NaOC2H5 → C4H8 + C2H5OH + NaCl but it is often carried out with an excess of NaOC2H5 to react with any water present in the reaction mixture that might reduce the yield. If the reaction shown was carried out with 6.83 g of C4H9Cl, how many grams of NaOC2H5 would be needed to have a 50 percent molar excess of the reactant

Answer 1

The mass of the sodium ethoxide is 7.4 g.

What is stoichiometry?

In stoichiometry, the coefficients in a balanced chemical equation represent the mole ratios between different substances. By understanding these ratios, one can determine the amount of reactants needed or products formed in a chemical reaction.

Number of moles of
`C_4H_9Cl` = 6.83 g /93 g/mol

= 0.073 moles

Since the reaction is 1: 1, to have a 50% excess we should have half the concentration of
`C_4H_9Cl` which is 0.073 moles + 0.0365 = 0.1095 moles

Mass of the
`NaOC_2H_5`= 0.1095 moles * 68 g/mol

= 7.4 g

Missing parts;

The following reaction is stoichiometric as written
`C4H_9Cl + NaOC_2H_5 -- > C_4H_8 + C_2H_5OH + NaCl`but it is often carried out with an excess of
`NaOC_2H_5`to react with any water present in the reaction mixture that might reduce the yield. If the reaction shown was carried out with 6.83 g of
`C_4H_9Cl`, how many grams of
`NaOC_2H_5`would be needed to have a 50 percent molar excess of the reactant?