3CS2 + 6 NaOH==> 2 Na2CS3 + Na2CO3 + 3 H2O
how many grams of Na2CS3 are produced in the reaction of 65 ml of liquid CS2(d=1.26 g/ml) and 1.56 mol NaOH?
This is a limiting reagent problem.
Convert 65 mL CS2 to moles. moles - grams/molar mass. Grams is obtained from the density data.
a. Using the coefficients in the balanced equation, convert moles CS2 to moles Na2CS3.
b. Same procedure, convert moles NaOH to moles Na2CS3.
c. It is likely that the answer for a and be will be different which means one of them is wrong. The correct answer, in limiting reagent problems, is ALWAYS the smaller one.
d. Convert moles from part c to grams. g = moles x molar mass.
Post your work if you get stuck.
Well, chemistry is a tough subject, but I'll give it a shot! Let's break it down step by step.
First, we need to find the volume of CS2 in liters, since the density is given in grams per milliliter. 65 ml divided by 1000 ml will give us the volume in liters:
65 ml / 1000 ml = 0.065 L
Next, we need to figure out the number of moles of CS2. To do that, we use the formula:
moles = volume (L) x density (g/ml)
moles = 0.065 L x 1.26 g/ml = 0.0823 mol CS2
According to the chemical equation, the ratio between CS2 and Na2CS3 is 3:2. So, if we have 0.0823 mol of CS2, we can calculate the moles of Na2CS3:
moles of Na2CS3 = (0.0823 mol CS2 x 2 mol Na2CS3) / 3 mol CS2
moles of Na2CS3 ≈ 0.0549 mol Na2CS3
Now comes the fun part - converting moles to grams. The molar mass of Na2CS3 is required for this step. From the chemical formula, we can calculate it:
molar mass of Na2CS3 = (2 x atomic mass of Na) + atomic mass of C + (3 x atomic mass of S)
molar mass of Na2CS3 = (2 x 23.0 g/mol) + 12.0 g/mol + (3 x 32.1 g/mol)
molar mass of Na2CS3 ≈ 186.3 g/mol
Finally, we can calculate the grams of Na2CS3 using the formula:
grams of Na2CS3 = moles of Na2CS3 x molar mass of Na2CS3
grams of Na2CS3 ≈ 0.0549 mol Na2CS3 x 186.3 g/mol ≈ 10.2 g Na2CS3
So, approximately 10.2 grams of Na2CS3 are produced in the reaction. Hope this helps lighten up the chemistry load!
To find out how many grams of Na2CS3 are produced in the reaction, we need to use stoichiometry and the given quantities of CS2 and NaOH.
Step 1: Calculate the volume of CS2 in liters.
65 ml x (1 L / 1000 ml) = 0.065 L
Step 2: Calculate the mass of CS2.
Mass = Volume x Density
Mass = 0.065 L x 1.26 g/mL = 0.0819 g
Step 3: Convert the mass of CS2 to moles using its molar mass.
Molar Mass of CS2 = (12.01 g/mol x 1) + (32.06 g/mol x 2) = 76.13 g/mol
Moles of CS2 = Mass / Molar Mass
Moles of CS2 = 0.0819 g / 76.13 g/mol ≈ 0.001076 mol
Step 4: Use the mole ratio between CS2 and Na2CS3 from the balanced equation to find the moles of Na2CS3.
From the balanced equation: 3CS2 + 6NaOH → 2Na2CS3 + Na2CO3 + 3H2O
The mole ratio between CS2 and Na2CS3 is 3:2.
Moles of Na2CS3 = (Moles of CS2) x (2/3)
Moles of Na2CS3 = 0.001076 mol x (2/3) ≈ 0.000717 mol
Step 5: Convert the moles of Na2CS3 to grams using its molar mass.
Molar Mass of Na2CS3 = (22.99 g/mol x 2) + (12.01 g/mol) + (32.06 g/mol x 3) = 119.16 g/mol
Mass of Na2CS3 = Moles of Na2CS3 x Molar Mass
Mass of Na2CS3 = 0.000717 mol x 119.16 g/mol ≈ 0.0855 g
Therefore, approximately 0.0855 grams of Na2CS3 are produced in the reaction.
To determine the number of grams of Na2CS3 produced in the reaction, follow these steps:
Step 1: Calculate the number of moles of CS2
Given that the density of CS2 is 1.26 g/mL and the volume is 65 mL, we can calculate the mass of CS2.
Mass = density × volume
Mass of CS2 = 1.26 g/mL × 65 mL = 81.9 g
Next, we need to find the number of moles of CS2. To do this, divide the mass of CS2 by its molar mass.
Molar mass of CS2 = (12.01 g/mol × 1) + (32.07 g/mol × 2) = 76.14 g/mol
Number of moles of CS2 = Mass of CS2 / Molar mass of CS2
Number of moles of CS2 = 81.9 g / 76.14 g/mol = 1.075 mol
Step 2: Calculate the limiting reagent
To determine the limiting reagent, compare the stoichiometric ratio of CS2 to NaOH, which is 3:6. We have 1.075 mol of CS2, so we need to convert this to moles of NaOH.
Based on the stoichiometry, 3 mol of CS2 react with 6 mol of NaOH.
Moles of NaOH = (1.075 mol CS2) × (6 mol NaOH / 3 mol CS2) = 2.15 mol NaOH
Since we have 1.56 mol of NaOH, which is less than the calculated 2.15 mol NaOH, NaOH is the limiting reagent.
Step 3: Calculate the moles of Na2CS3 formed
From the balanced equation, we can see that for every 3 mol of CS2, 2 mol of Na2CS3 are produced.
Therefore, the number of moles of Na2CS3 formed is calculated as follows:
Moles of Na2CS3 = (2 mol Na2CS3 / 3 mol CS2) × (1.075 mol CS2)
Moles of Na2CS3 = 0.7167 mol Na2CS3
Step 4: Convert moles of Na2CS3 to grams
Finally, use the molar mass of Na2CS3 to convert moles to grams.
Molar mass of Na2CS3 = (22.99 g/mol × 2) + (12.01 g/mol × 1) + (32.07 g/mol × 1) = 91.07 g/mol
Grams of Na2CS3 = Moles of Na2CS3 × Molar mass of Na2CS3
Grams of Na2CS3 = 0.7167 mol Na2CS3 × 91.07 g/mol ≈ 65.29 g
Therefore, approximately 65.29 grams of Na2CS3 are produced in the reaction.