a) How many moles of CaBr2 would be consumed when reacted with 53.0 ml of a...

  

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a) How many moles of CaBr2 would be consumed when reacted with 53.0 ml of a 2.00 M H2S04 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 H2504 is needed to produce this much HBr? d) A 532 mL solution of 0.133 M CAB12 is mixed with 603 mL of 0.111 M H2SO4. What mass of HBr could be produced under these conditions? CaBr2(ag) + H2S04(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 H2S04 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 H2504 is needed to produce this much HBr? d) A 532 mL solution of 0.133 M CAB12 is mixed with 603 mL of 0.111 M H2SO4. What mass of HBr could be produced under these conditions? CaBr2(ag) + H2S04(aq) → CaSO4(s) + 2HBr(g)

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a mol of H2SO4 MV 2 53103 0106 mol of H2S04 ratio is 11 so mol of ... View the full answer