H 427 ml of 0216 MHCX lege ?courseld 1489505180OpenVellumHMAC- 939530874426c701b
ID: 714147 • Letter: H
Question
H 427 ml of 0216 MHCX lege ?courseld 1489505180OpenVellumHMAC- 939530874426c701b39d1fc67dee67eas10001 CHEM 1212 e Home Homework 2 roblem 4.82-Enhanced-with Feedback 556 ml of BaCl, solution is neded to precipitate Express the molarity to three significant digits all he sullate ion in a 756 g sample of Na SO, what is the molarity of the solution? M- - Part D 427 mLof0216 M Ha soiton i, needed to neutraíze a sol ton of Ca(OH), how many grams oGOH,must be in the soldm? Express the mass in grams to three significant digits 8Explanation / Answer
Write down the balanced chemical equation for the reaction between BaCl2 and Na2SO4.
BaCl2 (aq) + Na2SO4 (aq) --------> BaSO4 (s) + 2 NaCl (aq)
As per the stoichiometric equation,
1 mole BaCl2 = 1 mole Na2SO4.
Determine the mole(s) of Na2SO4 corresponding to 756 mg Na2SO4.
Determine the molar mass of Na2SO4.
The atomic masses are:
Na = 22.989
S = 320.65
O = 15.999
Gram molar mass of Na2SO4 = (2*22.989 + 1*32.065 + 4*15.999) g/mol = 142.039 g/mol.
Mole(s) of Na2SO4 corresponding to 756 mg = (756 mg)*(1 g)/(1000 mg)*(1 mole)/(142.039 g)
= 0.005322 mole.
As per the stoichiometric equation,
0.005322 mole Na2SO4 = 0.005322 mole BaCl2.
Molar concentration of BaCl2 = (moles of BaCl2)/(volume of BaCl2 solution in L)
= (0.005322 mole)/[(55.6 mL)*(1 L)/(1000 mL)]
= 0.09572 mol/L
9.57*10-2 M (ans).
Part D
Write the balanced chemical equation for the reaction between Ca(OH)2 and HCl.
Ca(OH)2 (aq) + 2 HCl (aq) ---------> CaCl2 (aq) + 2 H2O (l)
As per the stoichiometric equation,
1 mole Ca(OH)2 = 2 moles HCl.
Mole(s) of HCl corresponding to 42.7 mL of 0.216 M HCl = (volume of HCl in L)*(molar concentration of HCl)
= (42.7 mL)*(1 L)/(1000 mL)*(0.216 M)
= 0.0092232 mole.
As per the stoichiometric equation,
moles Ca(OH)2 = (0.0092232 mole HCl)*(1 mole Ca(OH)2)/(2 moles HCl)
= 0.0046116 mole.
The atomic masses are:
Ca = 40.078
O = 15.999
H = 1.008
The molar mass of Ca(OH)2 = (1*40.078 + 2*15.999 + 2*1.008) g/mol = 74.092 g/mol.
Mass of Ca(OH)2 corresponding to 0.0046116 mole = (moles of Ca(OH)2)*(molar mass of Ca(OH)2)
= (0.0046116 mole)*(74.092 g/mol)
= 0.3417 g
0.342 g (ans).
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