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>
Answer: False
Explanation: Solutions can never be Heterogenous
I believe the answer is density
Break down in to tiny prices as the water hit the tree
The balanced equation for
Ca(OH)2 + H3PO4→ Ca3(PO4)2 + H2O is
3 Ca(OH)2 +2 H3PO4→ Ca3(Po4)2 + 6 H2O
3 moles of Ca(OH)2 reacted with 2 moles of H3PO4 to form 1 mole of Ca3(PO4)2 and 6 moles of H2O
