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.
To solve this problem, we need to calculate the molar masses of each substance involved in the reaction. Then, we'll use stoichiometry to determine the mass of carbon disulfide formed and the excess amount of reagent left unreacted.
(a) To calculate the mass of carbon disulfide formed:
1. Determine the number of moles of sulfur and carbon using their respective masses.
- Molar mass of sulfur (S): 32 g/mol
- Molar mass of carbon (C): 12 g/mol
Moles of sulfur = mass of sulfur / molar mass of sulfur
= 96 g / 32 g/mol
= 3 moles
Moles of carbon = mass of carbon / molar mass of carbon
= 36 g / 12 g/mol
= 3 moles
2. Since the balanced equation tells us that the ratio of C:S is 1:2, the limiting reactant is carbon because it forms fewer moles of carbon disulfide.
3. Using the stoichiometry of the balanced equation, we know that 1 mole of carbon reacts to produce 1 mole of carbon disulfide.
Moles of carbon disulfide formed = moles of carbon
= 3 moles
4. Calculate the mass of carbon disulfide formed using its molar mass.
- Molar mass of carbon disulfide (CS2): 76 g/mol
Mass of carbon disulfide formed = moles of carbon disulfide formed * molar mass of CS2
= 3 moles * 76 g/mol
= 228 g
Therefore, the mass of carbon disulfide formed is 228 grams.
(b) To calculate the excess amount of the reagent left unreacted:
1. Since carbon is the limiting reactant, we can determine the amount of sulfur that reacted with it.
Moles of sulfur reacted = moles of carbon / stoichiometric coefficient of carbon in the balanced equation
= 3 moles / 1
= 3 moles
2. Calculate the mass of sulfur reacted using its molar mass.
- Molar mass of sulfur (S): 32 g/mol
Mass of sulfur reacted = moles of sulfur reacted * molar mass of sulfur
= 3 moles * 32 g/mol
= 96 g
3. Calculate the excess amount of sulfur left unreacted by subtracting the mass of sulfur reacted from the initial mass of sulfur.
Excess amount of sulfur = Initial mass of sulfur - Mass of sulfur reacted
= 96 g - 96 g
= 0 g
Therefore, the excess amount of the reagent (sulfur) left unreacted is 0 grams.