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
The melting point would decrease
Two lobes.
The shape of four sp3 hybrid orbitals is part s and part p-each has two lobes but one lobe is much bigger (which is the lobe used for bonding)
Answer:
C,B,A
Explanation:
C has the most similar structure
B has the second most similar structure
A has the least similair structure
Cl⁻ + H₂O(g) = HCl(g) + OH⁻
w - it is percent chloride ions in solid sea salt (as a rule 55%)
m - it is the mass of sea salt
m(Cl⁻)=mw/100
m(Cl⁻)/M(Cl)=V(HCl)/V₀
mw/{100M(Cl)}=V(HCl)/V₀
m=100M(Cl)V(HCl)/{wV₀} (<span>the mass of solid sea salt)</span>
V₀=22.4 L/mol
M(Cl)=35.45 g/mol
for example:
V(HCl)= 1.0 L
w=55%
m=100×35.45×1.0/{55×22.4}=2.88 g
