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
Answer:
the equilibrium concentration of [PCl₅] is 3.64*10⁻³ M
Explanation:
for the reaction
PCl₅(g) → PCl₃(g) + Cl₂(g)
where
Kc= [PCl₃]*[Cl₂]/[PCl₅] = 2.0*10¹ M = 20 M
and [A] denote concentrations of A
if initially the mixture is pure PCl₅ , then it will dissociate according to the reaction and since always one mole of PCl₃(g) is generated with one mole of Cl₂(g) , the total number of moles of both at the end is the same → they have the same concentration → [PCl₃(g)] = [Cl₂]=0.27 M
therefore
Kc= [PCl₃]*[Cl₂]/[PCl₅] = 0.27 M* 0.27 M /[PCl₅] = 20 M
[PCl₅] = 0.27 M* 0.27 M / 20 M = 3.64*10⁻³ M
[PCl₅] = 3.64*10⁻³ M
the equilibrium concentration of [PCl₅] is 3.64*10⁻³ M
It is given that the person weighs 62 kg = 62,000 g
Natural abundances in mass percent are:
O = 65%
C = 18%
H = 10%
N = 3.0%
Ca = 1.6%
P = 1.2%
Corresponding weights of the elements are:
O = 65/100 * 62000 g = 40.30 * 10^3 g
C = 18/100 * 62000 g = 11.16 * 10^3 g
H = 10/100 * 62000 g = 62.00 * 10^2 g
N = 3.0/100 * 62000 g = 18.60 * 10^2 g
Ca = 1.6/100 * 62000 g = 9.92 * 10^2 g
P = 1.2/100 * 62000 g = 7.44 * 10^2 g
