What is the limiting reactant and the mass of [tex]$SO_2$[/tex] (in g) produced when 27.3 g of [tex]CH_3SH[/tex] reacts with 38.6 g of [tex]O_2[/tex]?

The question is asking about a limiting reactant problem which is a common concept in Chemistry. To solve this, we first have to write out and balance the chemical equation for the reaction, CH3SH + 3O2 -> CO2 + 2H2O + SO2. In this balanced equation, we see that 1 mole of CH3SH reacts with 3 moles of O2 to produce 1 mole of SO2.

Next, we convert the given masses of CH3SH and O2 into moles using their molar masses. 27.3g CH3SH / 48.1 g/mol = 0.567 mol CH3SH and 38.6g O2 / 32 g/mol = 1.206 mol O2. From these, you can see that CH3SH is the limiting reactant since we need 3 moles of O2 for every mole of CH3SH.

Therefore, the mass of SO2 produced will be determined by the stoichiometry of the reaction and the amount of the limiting reactant. So, the moles of SO2 produced would be the same as the moles of the limiting reactant, CH3SH, which is 0.567 moles. Converting this back into grams gives us 0.567 mol x 64 g/mol = 36.28 g of SO2 produced.

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