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 6.352 moles of CS2 and 8.958 moles of NaOH? Express your answer in the correct number of significant figures.

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

From the balanced chemical equation, we can see that the mole ratio of CS2 to NaOH is 3:6, or 1:2. This means that for every mole of CS2, we need 2 moles of NaOH.

If we have 6.352 moles of CS2, we would need 2 * 6.352 = 12.704 moles of NaOH. But we only have 8.958 moles of NaOH, so NaOH is the limiting reactant.

Next, we look at the mole ratio of NaOH to Na2CS3, which is 6:2, or 3:1. This means that for every 3 moles of NaOH, 1 mole of Na2CS3 is produced.

So, if we have 8.958 moles of NaOH, we can produce 8.958 / 3 = 2.986 moles of Na2CS3.

Finally, we need to convert moles of Na2CS3 to grams. The molar mass of Na2CS3 is approximately 158.18 g/mol.

So, 2.986 moles of Na2CS3 * 158.18 g/mol = 472.3 g of Na2CS3.

Therefore, 472.3 g of Na2CS3 can be produced from the reaction of 6.352 moles of CS2 and 8.958 moles of NaOH.