Question

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)

Answer 1

Step-by-step 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