25) Considering the limiting reactant concept, how many moles of copper() sulfid

Question

25) Considering the limiting reactant concept, how many moles of copper() sulfide are reaction of 2.00 mole of copper and 2.00 mole of sulfur? 2 Cu(s)S(s) Cu2s(s) A) 2.00 mol B) 1.00 mol C) 0.500 mol D) 1.50 mol E) none of the above 26) If the theoretical yield of LiOH in the reaction below is 22.9 g and the actual yield is 20.4 g, calculate the percent yield. Given: LbO+ HaO-2LiOH A) 82.0 % B)112% C) 89.1 % D) 45.0 % E) none of the above 27) What are the products from the following single-replacement reaction? Cu(s) ZnSO4(aq) A) Cu and ZnSO3 B) CuO and ZnsO3 C) Cu and ZnS04 D) CuO and ZnS04 E) no reaction 28) If the following ions Ba+2, C, Kt and S2 are placed in a test tube, the precipitate formed is A) KeS B) KCI C) Bas D BaC2 E) no precipitate 29) What are the coefficients for the following reaction when it is properly balanced? Cu(s)+AgNO(aq)-CuNOlag)+Ag(s) B) 2, 2, 1,1 D) 2, 3, 6,1 E) none of the above 30) What is the systematic name for BF3? A) boron fluoride B) boron (II) fluoride C) boron hexafluoride D) boron trifluoride E) none of the above 31) What is the empirical formula for a compound containing 1.000 g tin (Sn) and 0.320 g fluorine? A) SnFB) SnF10 C) SnF2 D)SnF4 E) Sn2F3

Explanation / Answer

25)

Answer

B) 1 mol

Explanation

2Cu(s) + S(s) - - - - - - > Cu2S(s)

stoichiometrically, 2 moles of Cu react with 1mole of S

given moles of Cu = 2

given moles of S = 2

So, S is excess reagent and Cu is limiting reagent

Stoichiometrically, 2moles of Cu give 1mole of Cu2S

Therefore

The answer is B) 1.00mol

26)

Answer

C) 89.1%

Explanation

Percent yield = (Actual yield/Theoreical yield) ×100

Percent yield = (20.4g/22.9g)×100 = 89.1%

27)

E) No reaction

28)

C) BaS

29)

C) 1,2,1,2

30)

B) boron(III)flouride

31)

Answer

C) SnF2

Explanation

No of moles of Sn =1g /118.71g/mol = 0.008423

No of moles of F = 0.320g/18.998g/mol = 0.01684

mole ratio of Sn:F is 1:2

Therefore

Empirical formula is SnF2

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