Q#2..96g of sulphur reacts with 36g of carbon according to the following equation and produces carbon disulphide
2S + C = CS2
(a) the mass of carbon disulphide
(b) the excess amount of the reagent in grams that is left unreacted.
Mols of,
S=(96/32)mol =3 mol
C=(36/12)mol= 3 mol
2S + C --> CS2
(3-3) (3-1.5) (+1.5) mol
a)=(3/2)mol*78g/mol
b) 1.5 mols of Carbon
=(3/2)mol*12g/mol
what is the ph of 50ml buffer solution which is 2M in CH3COOH and 2M in CH3CooNa?
1 Initial PH before the addition of acids and base?
2 What is the new PH after 2ml of 6.00M HCl is added to this buffer?
3 what is the new PH after 2.00ml of 6.00M NaoH is added to the original buffer?
To find the mass of carbon disulphide (CS2) produced and the excess amount of the reagent left unreacted, we need to determine the limiting reagent first. The limiting reagent is the reactant that is completely consumed and limits the amount of product formed.
Step 1: Calculate the number of moles of sulphur (S) and carbon (C) using their respective masses and molar masses.
Molar mass of sulphur (S): 32.06 g/mol
Molar mass of carbon (C): 12.01 g/mol
Number of moles of sulphur (S) = mass (g) / molar mass (g/mol)
Number of moles of sulphur (S) = 96 g / 32.06 g/mol = 2.996 mol (approximately 3 mol)
Number of moles of carbon (C) = mass (g) / molar mass (g/mol)
Number of moles of carbon (C) = 36 g / 12.01 g/mol = 2.998 mol (approximately 3 mol)
Step 2: Determine the stoichiometric ratio between the reactants.
From the balanced equation, the stoichiometric ratio between sulphur (S) and carbon disulphide (CS2) is 1:1, meaning one mole of sulphur reacts with one mole of carbon disulphide.
Step 3: Identify the limiting reagent.
To identify the limiting reagent, we compare the number of moles of each reactant to the stoichiometric ratio.
Both sulphur (S) and carbon (C) are present in the same number of moles (approximately 3 mol). Therefore, sulphur (S) is the limiting reagent since it will be completely consumed first.
Step 4: Calculate the mass of carbon disulphide (CS2) produced.
From the balanced equation, the stoichiometric ratio between sulphur (S) and carbon disulphide (CS2) is 1:1. This means that for every 1 mol of sulphur, 1 mol of carbon disulphide is produced.
Since sulphur (S) is the limiting reagent and completely reacts, the number of moles of carbon disulphide (CS2) formed will also be 3 mol.
Mass of carbon disulphide (CS2) = number of moles × molar mass
Mass of carbon disulphide (CS2) = 3 mol × (12.01 g/mol + 32.06 g/mol)
Mass of carbon disulphide (CS2) = 3 mol × 44.07 g/mol
Mass of carbon disulphide (CS2) = 132.21 g
Therefore, the mass of carbon disulphide (CS2) produced is 132.21 grams.
Step 5: Calculate the excess amount of the reagent left unreacted.
To calculate the excess amount of the reagent left unreacted, we need to determine the amount of the non-limiting reagent that was not consumed.
Since sulphur (S) is the limiting reagent, and the stoichiometry between sulphur (S) and carbon (C) is 2:1, we use this ratio to calculate the amount of carbon (C) required for complete reaction with the limiting amount of sulphur (S).
Number of moles of carbon (C) needed = 2 × number of moles of sulphur (S)
Number of moles of carbon (C) needed = 2 × 3 mol
Number of moles of carbon (C) needed = 6 mol
Excess amount of carbon (C) = number of moles of carbon (C) initially present - number of moles of carbon (C) needed
Excess amount of carbon (C) = (36 g / 12.01 g/mol) - 6 mol
Excess amount of carbon (C) = 3 mol
Mass of excess carbon (C) = number of moles × molar mass
Mass of excess carbon (C) = 3 mol × 12.01 g/mol
Mass of excess carbon (C) = 36.03 g
Therefore, the excess amount of the reagent (carbon) left unreacted is approximately 36.03 grams.