For the reaction given below, how many grams of Na2CS3 are produced in the reaction of 65.0 mL of liquid CS2 (d = 1.26 g/mL) and 1.56 mol NaOH?

To determine the number of grams of Na2CS3 produced in the reaction, we need to use the balanced chemical equation and convert the given quantities to the desired unit.

First, let's write the balanced chemical equation for the reaction:

2 NaOH + CS2 → Na2CS3 + H2O

According to the balanced chemical equation, we need 2 moles of NaOH to produce 1 mole of Na2CS3.

Given:
- Volume of liquid CS2 = 65.0 mL
- Density of CS2 = 1.26 g/mL
- 1.56 mol NaOH

To find the number of moles of CS2, we need to use the volume and density information:

Volume of CS2 = 65.0 mL
Density of CS2 = 1.26 g/mL

Mass of CS2 = Volume × Density
Mass of CS2 = 65.0 mL × 1.26 g/mL

Now, let's calculate the mass of CS2:

Mass of CS2 = 65.0 mL × 1.26 g/mL = 81.9 g

Next, we need to calculate the number of moles of CS2 using its molar mass, which is 76.13 g/mol:

Number of moles of CS2 = Mass of CS2 / Molar mass of CS2
Number of moles of CS2 = 81.9 g / 76.13 g/mol

Now, let's calculate the number of moles of Na2CS3 produced using the mole ratio from the balanced chemical equation:

Number of moles of Na2CS3 = (Number of moles of CS2) / 2
Number of moles of Na2CS3 = (81.9 g / 76.13 g/mol) / 2

Finally, we can convert the moles of Na2CS3 to grams using its molar mass, which is 163.21 g/mol:

Mass of Na2CS3 = Number of moles of Na2CS3 × Molar mass of Na2CS3
Mass of Na2CS3 = (81.9 g / 76.13 g/mol) / 2 × 163.21 g/mol

Therefore, to find how many grams of Na2CS3 are produced in the reaction, you would need to perform the calculations outlined above.