Answer:
58.9g of SO2 is produced
8g of oxygen remains unconsumed
Explanation:
The burning of Carbon disulfide (CS2) in oxygen. gives the reaction:
CS2 (g) + 3O2 (g) → CO2 (g) + 2SO2 (g)
Molar mass of CS2 = 76.139 g/mol
Molar mass of O2 = 15.99 g/mol
Molar mass of SO2 = 64.066 g/mol
Number of moles of CS2 = 35g/ 76.139 g/mol =0.46 moles
Number of moles of O2 = 30g/15.999 g/mol =1.88 moles
From the chemical reaction
1 mole of CS2 react with 3 moles of O2 to give 2 moles of SO2
Thus 0.46 moles of CS2 reacts to form 2× 0.46 = 0.92 moles of SO2
Mass of SO2 produced = 0.92×64.07 = 58.9g of SO2 is produced
thus 0.46 moles of CS2 reacts with 3 × 0.46 moles of O2 which is =1.38 moles of O2
Thus oxygen is the limiting reactant with 1.88 - 1.38 = 0.496~~0.5 mole remaining
Or 8g of oxygen
58.9g of SO2 is produced
oxygen is the limiting
Answer:
1. Phosphoric Acid
: Catalyst
2. Methyl Anthranilate
: Reactive
3. Sodium Nitrite
: Reactive
4. Diethyl Ether
: Solvent and reactant
5. Nitrogen
: Sub-product
Explanation:
The phosphoric acid is used as a catalyst for the reaction, the methyl anthranilate will react with the sodium nitrite to produce methyl salicylate, along with the diethyl ether and the nitrogen is a sub-product of the reaction.
Answer:
I think it would be C but don't quote me on it
Explanation:
It adds another obstacle for the current to go through
Answer:
The answer to your question is -2855 J
Explanation:
Reaction
2C₂H₆ + 7O₂ ⇒ 4CO₂ + 6H₂O
Formula
Heat of reaction = ΔHrxn = ΣΔHrxn products - ΣΔHrxn reactants
Substitution
ΔHrxn = { 4(-393.5) + 6(-241.8)} - {2(-84.7) + 7(0)}
ΔHrxn = {-1574 -1450.8} - {-169.4}
ΔHrxn = -3024.8 + 169.4
ΔHrxn = -2855.4 J
