When the value of Ksp = 3.83 x 10^-11 (should be given - missing in your Q)
So, according to the balanced equation of the reaction:
and by using ICE table:
Ag2CrO4(s) → 2Ag+ (Aq) + CrO4^2-(aq)
initial 0 0
change +2X +X
Equ 2X X
∴ Ksp = [Ag+]^2[CrO42-]
so by substitution:
∴ 3.83 x 10^-11 = (2X)^2* X
3.83 x 10^-11 = 4 X^3
∴X = 2.1 x 10^-4
∴[CrO42-] = X = 2.1 x 10^-4 M
[Ag+] = 2X = 2 * (2.1 x 10^-4)
= 4.2 x 10^-4 M
when we comparing with the actual concentration of [Ag+] and [CrO42-]
when moles Ag+ = molarity * volume
= 0.004 m * 0.005L
= 2 x 10^-5 moles
[Ag+] = moles / total volume
= 2 x 10^-5 / 0.01L
= 0.002 M
moles CrO42- = molarity * volume
= 0.0024 m * 0.005 L
= 1.2 x 10^-5 mol
∴[CrO42-] = moles / total volume
= (1.2 x 10^-5)mol / 0.01 L
= 0.0012 M
by comparing this values with the max concentration that is saturation in the solution
and when the 2 values of ions concentration are >>> than the max values o the concentrations that are will be saturated.
∴ the excess will precipitate out
The answer to the question is true.
First you need to find out the Limiting reactant (LR). convert both reactants to the same thing. Check that the chemical equation is balanced. Now use stoichiometr and remember at moles, multiply: need moles, divide2 g / 42g/mol= 0.0477 mol propane mass propane/ Molar Mass propane = moles propane4 g / 32 g/mol= 0.125 mol oxygen X (1 mol/ 5 mol) = 0.025 mol propane oxygen is the LRmass O2 / MM O2 X (mol propane / mol O2)0.025 mol X (3 mol / 1 mol ) = .075 mol CO20.075 mol X (12 + 2*16) g /mol = 3.6 g CO2 In one step:2 g / 42g/mol X (3 mol / 1 mol ) X 48 g/mol = 6.86 g CO24 g / 32 g/mol X (3 mol / 5 mol ) X 48 g/mol = 3.6 g CO2mass/ MM X coefficient ratio X MM (new)
