Question

CaBr2(aq) + H2SO4(aq) → CaSO4(s) + 2HBr(g)

a) How many moles of CaBr2 would be consumed when reacted with 53.0 mL of a 2.00 M H2SO4 solution?

b) If you consume added 15.00 mL of Calcium Bromide, and produce 0.023 moles of HBr, what was the original concentration of the calcium bromide?

c) If 10.32 g of HB are generated, what volume of 0.100 M H2SO4 is needed to produce this much HBr?

d) A 532 mL solution of 0.133 M CaBr2 is mixed with 603 mL of 0.111 M H2SO4 . What mass of HBr could be produced under these conditions?

Answer 1

a)

mol of H2SO4 = MV = 2 * 53*10^-3 = 0.106 mol of H2SO4

ratio is 1:1 so

mol of CaBR2 =  0.106 mol will be consumed

b)

original concentration:

mol of HBr = 0.023; mol of CaBr2 = 1/2*mol o fHBr = 1/2*0.023 = 0.0115 mol of CaBr2

V = 15 mL = 15*10^-3 L

[CaBr2] = m ol/V = 0.0115 / (15*10^-3) = 0.7666 M

c)

mol of Hbr = mass/MW = 10.32/80.91194 = 0.12754 mol of HBr2

then..

mol o fH2SO4 reuqired = 1/2*0.12754 = 0.06377 mol of H2SO4

M = mol/V

V = mol/M = 0.06377/0.100

V = 0.6377 Liter = 637.7 mL

d)

mmol of H2SO4 = MV = 0.111*603 = 66.933

mmol of CaBr2 = MV = 0.133*532 = 70.756

ratio is 1:1 , so H2SO4 is limiting

66.933 mmol of H2SO4 should produce = 2*66.933 = 133.866 mmol o fHBr

mass = mmol*MW = (133.866*10^-3)(80.91194 )= 10.83 g of HBr