Answer:
310.53 g of Cu.
Explanation:
The balanced equation for the reaction is given below:
CuSO₄ + Zn —> ZnSO₄ + Cu
Next, we shall determine the mass of CuSO₄ that reacted and the mass Cu produced from the balanced equation. This can be obtained as follow:
Molar mass of CuSO₄ = 63.5 + 32 + (16×4)
= 63.5 + 32 + 64
= 159.5 g/mol
Mass of CuSO₄ from the balanced equation = 1 × 159.5 = 159.5 g
Molar mass of Cu = 63.5 g/mol
Mass of Cu from the balanced equation = 1 × 63.5 = 63.5 g
Summary:
From the balanced equation above,
159.5 g of CuSO₄ reacted to produce 63.5 g of Cu.
Finally, we shall determine the mass of Cu produced by the reaction of 780 g of CuSO₄. This can be obtained as follow:
From the balanced equation above,
159.5 g of CuSO₄ reacted to produce 63.5 g of Cu.
Therefore, 780 g of CuSO₄ will react to produce = (780 × 63.5)/159.5 = 310.53 g of Cu.
Thus, 310.53 g of Cu were obtained from the reaction.
Your answer would be 0.00285 moles.
8798076 atoms are in a 0.3500 mol of gold
I don’t know about 1 But number 2 is AB + xH2O = AB.xH2O
Answer:
ZnS(s) ⇄ S²⁻(aq) + Zn²⁺(aq)
Explanation:
First, we will write the molecular equation, since it is easier to balance.
2 HBr(aq) + ZnS(s) ⇄ H₂S(aq) + ZnBr₂(aq)
In the full ionic equation we include all ions and molecular species.
2 H⁺(aq) + 2 Br⁻(aq) + ZnS(s) ⇄ 2 H⁺(aq) + S²⁻(aq) + Zn²⁺(aq) + 2 Br⁻(aq)
In the net ionic equation we include only the ions that participate in the reaction and the molecular species.
ZnS(s) ⇄ S²⁻(aq) + Zn²⁺(aq)
