Hey there!:
As per solubility rules inorganic compounds that are soluble are
:Ammonium (NH4⁺),potassium (K⁺), sodium (Na⁺) : All ammonium, potassium and sodium salts are soluble. Exceptions:some transition metal compounds.
Bromides (Br⁻), chlorides (Cl⁻) and iodides (I⁻): Most bromides are soluble. Exceptions: salts containing silver, lead, and mercury.
So Rb2Cl2 is expected to be soluble.
As per solubility rules inorganic compounds that are insoluble are : Carbonates (CO₃²⁻), chromates (CrO₄²⁻), phosphates (PO₄³⁻), silicates (SiO₄²⁻): All carbonates, chromates, phosphates and silicates are insoluble. Exceptions: those of ammonium, potassium and sodium.
So CaCO3 is insoluble and precipitates
Therefore:
The completed equation is :
Rb2CO3(aq) + CaCl2(aq) → CaCO3(s) + Rb2Cl2(aq)
Hope that helps!
Answer:
Mass of Ag produced = 64.6 g
Note: the question is, how many grams of Ag is produced from 19.0 g of Cu and 125 g of AgNO3
Explanation:
Equation of the reaction:
Cu + 2AgNO3 ---> 2Ag + Cu(NO3)2
From the equation above, 1 mole of Cu reacts with 2 moles of AgNO3 to produce 2 moles of Ag and 1 mole of Cu(NO3)2.
Molar mass of the reactants and products are; Cu = 63.5 g/mol, Ag = 108 g/mol, AgNO3 = 170 g/mol, Cu(NO3)2 = 187.5 g/mol
To determine, the limiting reactant;
63.5 g of Cu reacts with 170 * 2 g of AgNO3,
19 g of Cu will react with (340 * 19)/63.5 g of AgNO3 =101.7 g of AgNO3.
Since there are 125 g of AgNO3 available for reaction, it is in excess and Cu is the limiting reactant.
63.5 g of Cu reacts to produce 108 * 2 g of Ag,
19 g of Cu will react to produce (216 * 19)/63.5 g of Ag = 64.6 g of Ag.
Therefore mass of Ag produced = 64.6g
