Answer:
Final molarity of iodide ion C(I-) = 0.0143M
Explanation:
n = (m(FeI(2)))/(M(FeI(2))
Molar mass of FeI(3) = 55.85+(127 x 2) = 309.85g/mol
So n = 0.981/309.85 = 0.0031 mol
V(solution) = 150mL = 0.15L
C(AgNO3) = 35mM = 0.035M = 0.035m/L
n(AgNO3) = C(AgNO3) x V(solution)
= 0.035 x 0.15 = 0.00525 mol
(AgNO3) + FeI(3) = AgI(3) + FeNO3
So, n(FeI(3)) excess = 0.00525 - 0.0031 = 0.00215mol
C(I-) = C(FeI(3)) = [n(FeI(3)) excess]/ [V(solution)] = 0.00215/0.15 = 0.0143mol/L or 0.0143M
Answer:
98.6 g/mol.
Explanation:
Equation of the reaction
HX + NaOH--> NaX + H2O
Number of moles = molar concentration × volume
= 0.095 × 0.03
= 0.00285 moles
By stoichiometry, 1 mole of HX reacted with 1 mole of NaOH. Therefore, number of moles of HX = 0.00285 moles.
Molar mass = mass ÷ number of moles
= 0.281 ÷ 0.00285
= 98.6 g/mol.
Al
Explanation:
The limiting reactant will be Al:
4Al + 3O₂ → 2Al₂O₃
The limiting reactant is the reactant in short supply in a chemical reaction.
Given parameters:
Mass of Al = 30g Molar mass = 27g/mol
Number of moles =
= 
Number of moles of Al = 1.111 mole
Mass of O₂ = 30g, molar mass = 32g/mol
Number of moles =
= 0.94mol
In the reaction:
4 moles of Al reacted with 3 moles of O₂
1.11moles of Al will require
= 0.83mole to react
But we have been given 0.94mole of O₂. This is more than required.
Therefore O₂ is in excess and Al is the limiting reactant.
Learn more:
Limiting reagents brainly.com/question/6078553
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Ion know yo but good luck
Answer: 4.4
Explanation:
Thats what my computer says after i submit answer