Answer:
Explanation:
The I₂ is the common substance in the two equations.
(1) IO₃⁻ + 5I⁻ + 6H⁺ ⟶ 3I₂ + 3H₂O
{2) I₂ + 2S₂O₃²⁻ ⟶ 2I⁻ + S₄O₆²⁻
From Equation (1), the molar ratio of iodate to iodine is
From Equation (2), the molar ratio of iodine to thiosulfate is
Combining the two ratios, we get
Answer:
What is the absolute temperature of this gas sample when the pressure is ... The total mass of the water vapor is 0.252 g ... A silver spoon with a mass of 25.04 g at a temperature of 100.00 ... A 0.821 gram sample of pure NH F was treated with 25.0 mL of 1.00 M NaOH and heated to drive off the NH, a.
Explanation:
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
Answer:
I’m trying to do something similar to that
Explanation: