If 0.400 mol of pCl_5 reacts with 1.40 mole of water, how many of HCl are produc

Question

If 0.400 mol of pCl_5 reacts with 1.40 mole of water, how many of HCl are produced? [HCl] moles Consider the balanced reaction: MgP = 2 HBr rightarrow MgBr_2 + H_2O Formula weights (g/mol) MgO = 40.3 HBr = 80.91 MgBr_2 = 184.1 H2O = 18.02 When 18.0 grams of MgO and 32.4 grams of HBr react, What is the limiting reactant? How many grams of magnesium bromide [F.W. 184.1] are formed in the reaction? How many grams of water are formed in the reaction? How many grams of the excess reactant are left at the end of the reaction?

Explanation / Answer

MgO + 2HBr --> MgBr2 + H2O

a) Moles of MgO = mass of MgO / Molar mass of MgO

              = 18 g / 40.3 g/mol = 0.44665

HBr moles = mass of HBr / Molar mass of HBr

               = 32.4 g/ 80.91 g/mol = 0.4

as per reaction HBr moles needed =2 x MgO moles = 2 x 0.44665 =0.8933 but we have only 0.4 moles HBr

hence HBr is limiting reagent

b) MgBr2 moles formed = (1/2) HBr moles            ( from coeffients of balanced equation)

           = ( 1/2) x 0.4 = 0.2

MgBr2 mass = moles x molar mass of MgBr2 = 0.2 mol x 184.1 g/mol   = 36.82 g

c) H2O moles formed = ( 1/2) HBr moles = 0.2 ,

H2O mass = moles of H2O x molar mass of H2O = 0.2 mol x 18.02 g/mol = 3.604 g

d) MgO moles consumed = (1/2) HBr moles = 0.2

MgO moles left unreacted = initial MgO moles - MgO moles reacted = 0.44665-0.2 = 0.24665

MgO mass left unreacted = MgO moles x molar mass fo MgO

                = 0.24665 mol x 40.3 g/mol = 9.94 g

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