The balanced equation for the above reaction is as follows;
2S + 3O₂ --> 2SO₃
Stoichiometry of O₂ to SO₃ is 3:2
O₂ is the limiting reactant and S is provided in excess. since O₂ is the limiting reactant, the whole amount is consumed in the reaction and amount of product formed depends on amount of limiting reactant present.
Number of O₂ moles reacted- 4 g / 32 g/mol = 0.125 mol
3 mol of O₂ forms 2 mol of SO₃
therefore when 0.125 mol of O₂ reacts number of SO₃ moles - 2/3 x 0.125 mol
Number of SO₃ moles formed - 0.0833 mol
Answer is 4) 0.08 mol
From the stoichiometry of the combustion reaction, we can see that 7.4 L of oxygen is consumed.
<h3>What is combustion?</h3>
Combustion is a reaction in which a substance is burnt in oxygen. The equation of the reaction is; C4H10O(l) + 6O2 (g) → 4CO2 (g) + 5H2O(l)
We can obtain the number of moles of CO2 from;
PV = nRT
n = 1.02 atm * 7.15 L/0.082 atm LK-1mol-1 * (125 + 273) K
n = 7.29 /32.6
n = 0.22 moles
If 6 moles of oxygen produces 4 moles of CO2
x moles of oxygen produces 0.22 moles of CO2
x = 0.33 moles
1 mole of oxygen occupies 22.4 L
0.33 moles of oxygen occupies 0.33 moles * 22.4 L/ 1 mole
= 7.4 L of oxygen
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According to markovnikov's rule of the electrophilic addition to an alkene, the electrophile, usually a proton, is more likely to add to the less-substituted carbon in a double bond.
With additional substituents present in this configuration, the intermediate carbocation is stabilised by being located on the more-substituted carbon.
The nucleophile will then end up in a double bond on the more-substituted carbon in a reaction that follows Markovnikov's rule.The outcome of some addition reactions is described by Markovnikov's rule or Markownikoff's rule in organic chemistry. Vladimir Markovnikov, a Russian scientist, created the rule in 1870.
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Answer:
1.044 g sample contains only vitamin C (C6H8O6) and sucralose (C12H19Cl3O8). When the sample is dissolved in water to a total volume of 33.0 mL, the osmotic pressure of the solution is 3.69 atm at 295 K.
Answer:
1. 0.125 mole
2. 42.5 g
3. 0.61 mole
Explanation:
1. Determination of the number of mole of NaOH.
Mass of NaOH = 5 g
Molar mass of NaOH = 23 + 16 + 1
= 40 g/mol
Mole of NaOH =?
Mole = mass /molar mass
Mole of NaOH = 5/40
Mole NaOH = 0.125 mole
2. Determination of the mass of NH₃.
Mole of NH₃ = 2.5 moles
Molar mass of NH₃ = 14 + (3×1)
= 14 + 3
= 17 g/mol
Mass of NH₃ =?
Mass = mole × molar mass
Mass of NH₃ = 2.5 × 17
Mass of NH₃ = 42.5 g
3. Determination of the number of mole of Ca(NO₃)₂.
Mass of Ca(NO₃)₂ = 100 g
Molar mass of Ca(NO₃)₂ = 40 + 2[14 + (3×16)]
= 40 + 2[14 + 48]
= 40 + 2[62]
= 40 + 124
= 164 g/mol
Mole of Ca(NO₃)₂ =?
Mole = mass /molar mass
Mole of Ca(NO₃)₂ = 100 / 164
Mole of Ca(NO₃)₂ = 0.61 mole
