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chm 101 quiz #5 limiting reactant problem
1. (10 points) in the raschig process, ammonia, nh3, is oxidized with sodium hypochlorite
(formed from naoh and cl2) to produce hydrazine.
2 naoh(s) + 1 cl2 (g) + 2 nh3(g) → 1 n2h4 (1) + 2 nacl (s) + 2 h2o (i)
a) if 25.0 grams of naoh reacted with 30.00 grams of cl2, in the presence of excess nh3,
which is the limiting reactant, and what is the theoretical yield of n2h4? .
b) if 8.00 grams of n2h4 is recovered, what is the percent yield?

Ineed someone to see if my answers are correct! if any are wrong let me know what the correct answers would be and how to get that answer! 1. how many moles of sodium chloride are in 28 grams od nacl? a. 265 mole naclb. 856 mole naclc. 479 mole of nacld. 1.2 mole nacl < my choice2. 734 grams of lithium sulfate (li2so4) are dissolved to make 2500 ml of solution what is rhe molaratiy? a. 2.67 mb. 4.56 mc. 3.89 m < my choiced. 1.78 m3. how many grams of cacl2 would be dissolved in 3.0 l of a 0.50 m solution of cacl2? a. 250 g cacl2 b. 166.5 g cacl2c. 113.65 g cacl2d. 98 g cacl2 < my choice4. suppose you had 58.44 g of nacl and you dissolved it in exactly 2.00 liters. the molarity if the solution would be 0.5 mtrue < my choicefalse 5. i would need 22g of naoh to make a 3.0 m solution using 250 ml of solvent.true < my choicefalse6. identify the solute: you have a .0195 m solution made from using 6.5 g of solute and 3 l of solvent. identify the solute by solving for molar weight.a. the solute is nacl because the molar weight is 58.43 g/mol < my choiceb. the solute is h2so4 because the molar weight is 98.06 g/molc. the solute is cacl2 because the molar weight is 111.11 g/mol