How many grams of CuSO_4 middot 5 H_2 O(s) are required to produce 500 millilite

Question

How many grams of CuSO_4 middot 5 H_2 O(s) are required to produce 500 milliliters of an aqueous solution that contains 250-milligrams Cu^2+ (aq) per liter? What volume of 1.250 M HCl(aq) is required to react with 11.78 grams of Na_2 CO_3(s)? The equation for the reaction is Na_2 CO_3 (s) + 2 HCl(aq) rightarrow 2 NaCl (aq) + H_2 O(l) + CO_2 (g) What volume of 0.865 M CaCl_2 (aq) is required to precipitate all the Ag^+(aq) from 35.0 milliliters of 0.500 M AgNO_3 (aq)? How many grams of AgCl(s) will precipitate? A 6.076-gram sample of succinic acid,

Explanation / Answer

The balance reaction between CaCl2 and Ag+ is as follows:

2Cl-(aq) + 2Ag+(aq) 2AgCl(s)

First calculate the moles of Ag+ ions as follows:

0.035 L * 0.500 moles / L= 0.0175 Moles Ag+

Now calculate the moles of Cl- as follows:

0.0175 Moles Ag+ *2.0 Moles of Cl-/2.0 Moles Ag+

0.0175 Moles Cl-

One mole of CaCl2 gives 2 moles Cl-.

0.0175 Moles Cl- * 1.0 moles CaCl2/ 2 moles Cl-

=8.75*10^-3 moles CaCl2

Volume of CaCl2 = number of moles / molarity

= 8.75*10^-3 moles CaCl2/ 0.865 M or moles / L

= 0.010 L or 10.0 mL

Moles of AgCl:

0.0175 Moles Cl- *2.0 moles of AgCl /2.0 Moles Cl-

=0.0175 Moles AgCl

Amount of AgCl:

Number of moles * molar mass

= 0.0175 Moles AgCl*143.32 g/mol

= 2.5081 g AgCl

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