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When 1.20 moles of MgO reacts as completely as possible with 0.900 moles of NaCl, the total number of moles of reaction products is:

MgO (s) + 2 NaCl (aq) → MgCl2 (aq) + Na2O (aq)

Respuesta :

linahamdan1980

Answer:

Explanation:

limiting reactant is NaCl it is consumed completely)

calculate n for each product

  n NaCl (mole )                 n MgCl2(mole )

    2                                                1

    0.9                                              x

        x = 0.45 mole

 n NaCl                          n Na2O

  2                                      1

   0.900                    x

 x = 0.45 mole

  the total number of moles of reaction products is: 0.45 + 0.45= 0.9mol

 

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