CS2 and NaOH react according to the equation below.3CS2 + 6NaOH ⟶ 2Na2CS3 + Na2CO3 + 3H2OHow many grams of Na2CS3 can be produced from the reaction of 92.50 moles of CS2 and 62.80 moles of NaOH? Express your answer in the correct number of significant figures.

To solve this problem, we need to first determine which reactant is the limiting reactant, as it will determine the maximum amount of Na2CS3 that can be produced.

1. Calculate the number of moles of Na2CS3 that can be produced from each reactant:

From the balanced chemical equation, we know that 3 moles of CS2 react with 6 moles of NaOH to produce 2 moles of Na2CS3.

So, from 92.50 moles of CS2, we can produce (92.50 moles CS2 x 2 moles Na2CS3 / 3 moles CS2) = 61.67 moles of Na2CS3.

From 62.80 moles of NaOH, we can produce (62.80 moles NaOH x 2 moles Na2CS3 / 6 moles NaOH) = 20.93 moles of Na2CS3.

2. Determine the limiting reactant:

The limiting reactant is the one that produces the least amount of product, which in this case is NaOH. Therefore, the maximum amount of Na2CS3 that can be produced is 20.93 moles.

3. Convert moles of Na2CS3 to grams:

The molar mass of Na2CS3 is approximately 158.18 g/mol.

So, the mass of Na2CS3 that can be produced is (20.93 moles Na2CS3 x 158.18 g/mol) = 3313 g or 3.31 kg.

Therefore, 3.31 kg of Na2CS3 can be produced from the reaction of 92.50 moles of CS2 and 62.80 moles of NaOH.