I need to show me the steps of the correct answer for 15 and 16 please What mass
ID: 1022091 • Letter: I
Question
I need to show me the steps of the correct answer for 15 and 16 please
What mass (un g) is needed to make 50.0 mL of a 0.125 m solution of chloride ion using calcium chloride? Molar mass (caCl_2) = 110.98 g mol^-1 0.277 g 0.347 g 0.694 g 1.39 g An excess of sodium sulfate reacts with 25.00 mL of a 0.138 M solution of silver nitrate. What is the percent yield when the precipitate is dried and weighed and found to have a mass of 0.495 g? Na_2SO_4(aq) + 2AgN0_3(aq) rightarrow Ag_2SO_4(s) + 2NaNO_3 (aq) 23.0% 27.9% 46& 92.0%Explanation / Answer
Formula of calcium chloride is used to find the molar ratio between calcium chloride and chloride ion. Using volume and molarity of the chloride ion, moles of it are calculated. Moles of chloride ion are used to get moles of calcium chloride.
Using given molar mass is use to find mass of it in g.
Moles of Cl- = volume in L x molarity = 0.050 L x 0.125 M = 0.00625 mol
Mole ratio between chloride ion : calcium chloride is 2 : 1
Therefore,
n Calcium chloride = n Cl- x 1 mol Calcium chloride / 2 mol Cl-
= 0.00625 mol Cl- x 1 mol CaCl2 / 2 mol Cl-
= 0.003125 mol CaCl2
Mass of CaCl2 = 0.003125 mol CaCl2 x 110.98 g /mol
= 0.347 g Calcium chloride
16.
Given reaction is used to find mole ratio between limiting reactant and the Ag2SO4 ( ppt)
Since the sodium phosphate is used in excess, silver nitrate would be the limiting reactant.
Mole ratio between Silver nitrate: silver sulfate is 2 : 1
Moles of silver sulfate formed in this reaction are calculated using the mole ratio.
n Ag2SO4 = moles of AgNO3 x 1 mol Ag2SO4 / 2 mol AgNO3
since moles of AgNO3 = volume in L x molarity,
n Ag2SO4 = 0.025 L x 0.138 M x 1 mol Ag2SO4 / 2 mol AgNO3
=0.00345 mol
Moles CaCl2 = 0.00345 x 1 / 2 ] moles
= 0.001725 moles
Mass in g = 0.001725 mol x molar mass = 0.001725 mol x 311.81 g /mol = 0.537872 g ( Theoretical yield)
We know percent yield = ( Actual yield / theoretical yield) x 100 %
= ( 0.495 / 0.5378) x 100
=92.0 %