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Question

1) A flask contains 81.9 mL of 0.150 M Ca(OH)2. How many milliliters of 0.350 M Na2CO3 are required to react completely with the calcium hydroxide in the following reaction?
Na2CO3(aq) _ Ca(OH)2(aq) --> CaCO3(s) + 2NaOH(aq)

2) A 1.345 g sample of a compound of barium and oxygen was dissolved in hydrochloric acid to give a solution of barium ion, which was then precipitated with an excess of potassium chromate to give 2.012 g of barium chromate, BaCrO4. What is the formula of the compound?

Answer 1

1) Let Ca(OH)2 be called CH. Let Na2CO3 be called SC.

81.9mLCH x 0.150molCH/1000mLCH x 1molSC/1molCH x 1000mLSC/0.350molSC = 35.1mL Na2CO3 solution

2) Atomic weights: Ba=137 Cr=52 O=16 BaCrO4=253

2.012gBaCrO4 x 1molBaCrO4/253gBaCrO4 x 1molBa++/1molBaCrO4 x 137gBa++/1molBa++ = 1.090g Ba++

1.345gsample - 1.090gBa = 0.255gO

1.090gBa x 1molBa/137gBa = 0.00796 mole Ba

0.255gO x 1molO/16gO = 0.016 mole O

The molO/molBa ratio = 2. So the formula is BaO2. This is barium peroxide. The usual oxide of barium is BaO. But the answer BaO2 is the right one.