Answer:
25.53mL of 0.200 M FeCl₃ are needed to produce 0.345g of Fe₂S₃
Explanation:
Based on the reaction of the problem, 1 mole of Fe₂S₃ is produced from 2 moles of FeCl₃.
0.345g of Fe₂S₃ are (Molar mass: 207.9g/mol):
0.345g of Fe₂S₃ ₓ (1 mol / 207.9g) = <em>1.6595x10⁻³ moles Fe₂S₃</em>
Moles of Fe needed to produce these moles of Fe₂S₃ are:
1.6595x10⁻³ moles Fe₂S₃ ₓ ( 2 moles FeCl₃ / 1 mole Fe₂S₃) =
<em>3.3189x10⁻³ moles of FeCl₃</em>
As the percent yield of the reaction is 65.0%, the moles of FeCl₃ you need to add are:
3.3189x10⁻³ moles of FeCl₃ ₓ (100.0% / 65.0%) = <em>5.106x10⁻³ moles of FeCl₃</em>
A solution 0.200M contains 0.200 moles per L. Volume to obtain 5.106x10⁻³ moles is:
<em>5.106x10⁻³ moles of FeCl₃ ₓ ( 1L / 0.200mol) = 0.02553L = </em>
<h3>25.53mL of 0.200 M FeCl₃ are needed to produce 0.345g of Fe₂S₃</h3>
The number of dots represents the amount of valence electrons, which is the same as the last digit of the elements group number in ptof.
2AgNO3 + Zn --> Zn(NO3)2 + 2Ag
Its a redox reaction, Zn is oxidised (0 --> 2+), Ag is reduced (1+ --> 0)
Answer:
51.53 grams .
Explanation:
Na₃PO₄ ⇄ 3Na⁺¹ + PO₄⁻³ .
1 mole 3 mole
725 mL of 1.3 M Na⁺ ions
= .725 x 1.3 moles of Na⁺ ions
= .9425 moles
3 mole of Na⁺ is formed by 1 mole of Na₃PO₄
.9425 mole of Na⁺ is formed by .9425/3 mole of Na₃PO₄
Na₃PO₄ needed = .9425/3 moles = .3142 moles
Molecular weight of Na₃PO₄ = 164
grams of Na₃PO₄ needed = .3142 x 164 = 51.53 grams .
