55.0 mL of 0.500 M lead(II) nitrate (FW = 331.2) is reacted with excess potassium sulfide aqueous solution. Calculate the quantity of lead(II) sulfide (FW = 239.3) precipitated in this reaction. Pb(NO3)2(aq) + K2S(aq)→ PbS(s) + 2 KNO3(aq) O 0.658 g O 6.58 g O 7.61 8 O 3.29 g

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Chapter3: Chemical Reactions
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U 0.00213 11
O 0.00417 M
Question 19
1 p
55.0 mL of 0.500 M lead(II) nitrate (FW = 331.2) is reacted with excess potassium sulfide aqueous solution. Calculate the quantity of
lead(II) sulfide (FW = 239.3) precipitated in this reaction.
Pb(NO3)2(aq) + K2S(aq) → PbS(s) + 2 KNO3(aq)
O 0.658 g
O 6.58 g
O 7.61 g
O 3.29 g
Question 20.
1 pts
80
888
F3
F4
%24
Transcribed Image Text:U 0.00213 11 O 0.00417 M Question 19 1 p 55.0 mL of 0.500 M lead(II) nitrate (FW = 331.2) is reacted with excess potassium sulfide aqueous solution. Calculate the quantity of lead(II) sulfide (FW = 239.3) precipitated in this reaction. Pb(NO3)2(aq) + K2S(aq) → PbS(s) + 2 KNO3(aq) O 0.658 g O 6.58 g O 7.61 g O 3.29 g Question 20. 1 pts 80 888 F3 F4 %24
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