what is maximum number of grams of vermillion you can make frim reactiom of 60.0

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what is maximum number of grams of vermillion you can make frim reactiom of 60.0 ml of a 0.150 M aqueous solution of Hg(NO3)2 with an excess of sodium sulfide nod/smartwork/view.php?id-5494208isStudent-1 D Bing Gmail (3) G Gmail (2) G Gmail Bank of America Suggested Sites C Vermillion is a very rare and expensive solid natural pigment known as Chinese red; it is the pigment used to print the name chops on works of art. It is mercury (l) sulfide (HgS), and it is also very insoluble in water. What is the maximum number of grams of vermillion you can make from the reaction of 60.0 mL of a 0.150 M aqueous solution of mercury (II) nitrate [Hg(NO3)2] with an excess of a concentrated aqueous solution of sodium sulfide (Na S)? Number Tool Check Answer ? View S

Explanation / Answer

Moles of mercury(II)nitrate = molarity*volume of solution in litres = 0.15*0.06 = 0.009

Since 1 mole of mercury(II)nitrate can produce 1 mole of vermillion on complete reaction

Thus, moles of vermillion produced = moles of mercury(II)nitrate = 0.009

mass of vermillion that can be produced = moles*molar mass = 0.009*232.66 = 2.094 g

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