A) A chemist wants to make 6.5 L of a 0.310 M CaCl 2 . What mass of CaCl 2 (in g

Question

A) A chemist wants to make 6.5 L of a 0.310 M CaCl2. What mass of CaCl2 (in g) should the chemist use?

B) What mass of silver chloride cna be produced from 1.00 L of a 0.104 M solution of silver nitrate? The reaction described required 3.79 L of magnesium chloride. What is the concentration of this magnesium chloride solution? Express your answer with the appropriate units

C) What volume (in mL) of a 0.300 M HNO3 solution will completely react with 26.6 mL of a 0.117 M Na2CO3 soltuion according to this balanced equation? Na2CO3(aq)+2 HNO3(aq)----> 2NaNO3(aq)+CO2(g)+H2O(l)

D) Consider the following precipitation reaction: 2 Na3PO4(aq)+3CuCl2(aq)---->Cu3(PO4)2(s)+6 NaCl(aq). What volume of 0.190 M Na3PO4 solution is neccessary to completely react with 93.3 mL of a 0.110 M CuCl2

Explanation / Answer

A) Number of moles of CaCl2 = 6.5*0.31 = 2.015

Thus weight = 111*2.015 = 223.66 g

B) reaction missing

C) The reaction occurs in the ratio 1:2 . Let the volume be V mL. Thus

2*26.6*0.117 = 0.3*V

Thus V = 20.748 mL

D) The reaction occurs in the ratio 2:3. Let the volume be V mL. Thus

93.3*0.11*2 = 0.19*V*3

Thus, V = 36.01 mL

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