Nickel(II) sulfate is prepared by adding an excess of nickel(II) carbonate to 0.010 mol of dilute sulfuric acid. NiCO3(s)   +   H2SO4(aq) -> NiSO4(aq)   +   H2O(l)   +  ...

Nickel(II) sulfate is prepared by adding an excess of nickel(II) carbonate to 0.010 mol of dilute sulfuric acid.
 NiCO3(s)   +   H2SO4(aq) -> NiSO4(aq)   +   H2O(l)   +   CO2(g)
Solid nickel(II) sulfate crystals are produced with a 20% yield.  How many moles of nickel(II) sulfate crystals are obtained?
  A) 0.001
  B) 0.002
  C) 0.010
  D) 0.050

 

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llltkl | College Teacher | (Level 3) Valedictorian

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The reaction is:  
 NiCO3(s)   +   H2SO4(aq) -> NiSO4(aq)   +   H2O(l)   +   CO2(g)

Nickel(II) carbonate is taken in excess, 0.010 mol H2SO4 is taken.

So, H2SO4 is the limiting reagent. Yield of NiSO4 is 20%.

Considering  the reaction stoichiometry and yield percent, it can be said that,

 1.0 mole H2SO4 should yield 0.2 mole NiSO4.

So, 0.01 mole H2SO4 produces 0.2*0.01 = 0.002 mole NiSO4.

Therefore the correct answer is option B), 0.002 moles.

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