SHON YOUR WORK (12 points, 4 points each) Detail step by step work and unit must

Question

SHON YOUR WORK (12 points, 4 points each) Detail step by step work and unit must be shown to get the full point. 1. An unknown compound with a molar mass of 223.94 g/mol consists of 32.18%,C,4,50% H, and 63.32%-CL Find the molecular formula for the compound. CHCI B CeH1oCl C) C3HsCl2 D) CoHhsCl 2. Hydrogen chloride gas can be prepared by the following reaction: 2NaCl(s) + H2SOdaq) ? 2HCl(g) + Na2SO4(s) How many grams of HCl can be prepared from 2.00 mol H2S04 and 2.56 mol NaCI? Which is limiting reagent? A) 7.30 g B) 93.3 g C) 146 g D) 150 g E) 196 g

Explanation / Answer

1. Molar mass = 223.94 g/mol

Percentage of carbon = 32.18%

Contribution of carbon to molar mass = 223.94*32.18/100 which is approximately equal to 72.

Molar mass of one carbon is 12. So there must be 72/12=6 carbons in the unknown compound.

Option B and E seem to satisfy this condition.

Now for chloride, its contribution = 63.32*223.94/100 = 141.8

Molar mass of one chlorine is 35.5 so there must be 141.8/35.5 that means 4 chlorine atoms.

This is only satisfied by option b.

Now check, C6H10Cl4 molar mass = (12*6)+10+(35.5*4)=224 which is correct within limits of error in molar mass.

2. 2 moles of NaCl react with 1 mol of H2SO4

2.56 mol NaCl reacts with 1.28 mol of H2SO4

Out of 2 mol, 0.72 mol is left unreacted. Therefore H2SO4 is in excess and NaCl is the limiting reagent.

2mol of NaCl give 2mol of HCl.

2.56mol of NaCl give 2.56mol of HCl.

Mass of HCl = 36.5*2.56 = 93.4

Option b is the answer.

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