Answer:
A = 2A + 3B → 5C
Explanation:
The two molecule of A and three molecules of B will react to form the five molecules of C.
2A + 3B → 5C
Other options are incorrect because,
B = A₂ + B₃ → C₅
in this reaction one molecule of A₂ and one molecule of B₃ combine to form one molecule of C₅.
C = 2A + 5B → 3C
in this reaction two molecules of A and five molecules of B combine to form three molecule of C.
D = A₂ + B₃ → C₃
in this reaction one molecule of A₂ and one molecule of B₃ combine to from one molecule of C₃.
Answer:
Part 1: 7.42 mL; Part 2: 3Cu²⁺(aq) + 2PO₄³⁻(aq) ⟶ 2Cu₃(PO₄)₂(s)
Explanation:
Part 1. Volume of reactant
(a) Balanced chemical equation.
![\rm 2Na_{3}PO_{4} + 3CuCl_{2} \longrightarrow Cu_{3}(PO_{4})_{2} + 6NaCl](https://tex.z-dn.net/?f=%5Crm%202Na_%7B3%7DPO_%7B4%7D%20%2B%203CuCl_%7B2%7D%20%5Clongrightarrow%20Cu_%7B3%7D%28PO_%7B4%7D%29_%7B2%7D%20%2B%206NaCl)
(b) Moles of CuCl₂
![\text{Moles of CuCl}_{2} =\text{ 16.7 mL CuCl}_{2} \times \dfrac{\text{0.200 mmol CCl}_{2}}{\text{1 mL CuCl}_{2}} = \text{3.340 mmol CuCl}_{2}](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20CuCl%7D_%7B2%7D%20%3D%5Ctext%7B%2016.7%20mL%20CuCl%7D_%7B2%7D%20%5Ctimes%20%5Cdfrac%7B%5Ctext%7B0.200%20mmol%20CCl%7D_%7B2%7D%7D%7B%5Ctext%7B1%20mL%20CuCl%7D_%7B2%7D%7D%20%3D%20%20%5Ctext%7B3.340%20mmol%20CuCl%7D_%7B2%7D)
(c) Moles of Na₃PO₄
The molar ratio is 2 mmol Na₃PO₄:3 mmol CuCl₂
![\text{Moles of Na$_{3}$PO}_{4} = \text{3.340 mmol CuCl}_{2} \times \dfrac{\text{2 mmol Na$_{3}$PO}_{4}}{\text{3 mmol CuCl}_{2}} =\text{2.227 mmol Na$_{3}$PO}_{4}](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20Na%24_%7B3%7D%24PO%7D_%7B4%7D%20%3D%20%20%5Ctext%7B3.340%20mmol%20CuCl%7D_%7B2%7D%20%5Ctimes%20%5Cdfrac%7B%5Ctext%7B2%20mmol%20Na%24_%7B3%7D%24PO%7D_%7B4%7D%7D%7B%5Ctext%7B3%20mmol%20CuCl%7D_%7B2%7D%7D%20%3D%5Ctext%7B2.227%20mmol%20Na%24_%7B3%7D%24PO%7D_%7B4%7D)
(d) Volume of Na₃PO₄
![V = \text{2.227 mmol Na$_{3}$PO}_{4}\times \dfrac{\text{1 mL Na$_{3}$PO}_{4}}{\text{0.300 mmol Na$_{3}$PO}_{4}} = \text{7.42 mL Na$_{3}$PO}_{4} \\\\\text{The reaction requires $\large \boxed{\textbf{7.42 mL Na$_{3}$PO}_{4}}$}](https://tex.z-dn.net/?f=V%20%3D%20%5Ctext%7B2.227%20mmol%20Na%24_%7B3%7D%24PO%7D_%7B4%7D%5Ctimes%20%5Cdfrac%7B%5Ctext%7B1%20mL%20Na%24_%7B3%7D%24PO%7D_%7B4%7D%7D%7B%5Ctext%7B0.300%20mmol%20Na%24_%7B3%7D%24PO%7D_%7B4%7D%7D%20%3D%20%5Ctext%7B7.42%20mL%20Na%24_%7B3%7D%24PO%7D_%7B4%7D%20%5C%5C%5C%5C%5Ctext%7BThe%20reaction%20requires%20%24%5Clarge%20%5Cboxed%7B%5Ctextbf%7B7.42%20mL%20Na%24_%7B3%7D%24PO%7D_%7B4%7D%7D%24%7D)
Part 2. Net ionic equation
(a) Molecular equation
![\rm 2Na_{3}PO_{4}(\text{aq}) + 3CuCl_{2}(\text{aq}) \longrightarrow Cu_{3}(PO_{4})_{2}(\text{s}) + 6NaCl(\text{aq})](https://tex.z-dn.net/?f=%5Crm%202Na_%7B3%7DPO_%7B4%7D%28%5Ctext%7Baq%7D%29%20%2B%203CuCl_%7B2%7D%28%5Ctext%7Baq%7D%29%20%5Clongrightarrow%20Cu_%7B3%7D%28PO_%7B4%7D%29_%7B2%7D%28%5Ctext%7Bs%7D%29%20%2B%206NaCl%28%5Ctext%7Baq%7D%29)
(b) Ionic equation
You write molecular formulas for the solids, and you write the soluble ionic substances as ions.
According to the solubility rules, metal phosphates are insoluble.
6Na⁺(aq) + 2PO₄³⁻(aq) + 3Cu²⁺(aq) + 6Cl⁻(aq) ⟶ Cu₃(PO₄)₂(s) + 6Na⁺(aq) + 6Cl⁻(aq)
(c) Net ionic equation
To get the net ionic equation, you cancel the ions that appear on each side of the ionic equation.
<u>6Na⁺(aq)</u> + 2PO₄³⁻(aq) + 3Cu²⁺(aq) + <u>6Cl⁻(aq)</u> ⟶ Cu₃(PO₄)₂(s) + <u>6Na⁺(aq)</u> + <u>6Cl⁻(aq)</u>
The net ionic equation is
3Cu²⁺(aq) + 2PO₄³⁻(aq) ⟶ Cu₃(PO₄)₂(s)
Answer:
beacuse water is same every where because it is the combination of h2 +o2 h2o which doesn't change while it is different
Explanation:
ok