28. Determine the theoretical yield of HCl if 30.0 g of BCl3 and 32.5 g of H2O a

Question

28. Determine the theoretical yield of HCl if 30.0 g of BCl3 and 32.5 g of H2O are reacted according to the following balanced reaction. A possibly useful molar mass is BCl3 = 117.2 g/mol. BCl3(g) + 3H2O(l) 3H3BO3(s) + 3HCl(g)

29. Which should give the most vigorous reaction when dropped in water?

(A). Mg (B). C (C). Na (D). Ca (E). U

30. Lithium and nitrogen react to produce lithium nitride: 6Li(s) + N2(g) --> 2Li3N(s) How many moles of N2 are needed to react with 0.250 mol of lithium?

(A). 12.0 (B). 1.45 (C). 2.99 (D). 1.50 (E). No correct answer is given

Explanation / Answer

28)no. of moles of BCl3 for 30g =30/117.2=0.25moles;for water moles=32.5/18=1.8moles

BCl3(g) + 3H2O(l) H3BO3(s) + 3HCl(g)

from balanced equation 1mole of BCl3 react with 3moles of water

for 0.25 moles of BCl3=?moles of water is required

=0.25*3=0.75moles

wateris in excess so the limiting reagent is BCl3

if 1mole of BCl3 gives 3 moles of HCL from equation

0.25 moles of BCl3=?

the no. of moles of HCl produced=0.25*3=0.75moles=0.75*36.5=27.375g

29) sodium(Na)

30)0.25/6=0.0416moles

Related Questions

Navigate

• Browse (All)
• Browse 2
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.