How many grams of mgcl2 will be produced from 10.5g of mg(oh)2 and 41g of hcl?

The number  of grams  of MgCl2 that   will  be  produced  from 10.5 g Mg(OH)2 and 41 of HCl  is  17.154 grams

calculation

write the  equation  for reaction
 Mg(OH)2  + HCl →  MgCl2  +2H2O

find the moles of  each compound  used

moles=mass /molar mass

moles of  Mg(OH)2 =  10.5g/ 58.3 g/mol =  0.180  moles
moles  of  HCl =  41 g/ 36.5g/mol = 1.123  moles

by  use of of mole ratio between  Mg(OH)2 to  MgCl2  which  is 1:1  moles of MgCl2 is also   0.180  moles

by  use   of mole ratio  between  HCl  to  MgCl2 which is 2:1 the moles of MgCl2  = 1.123  x1/2 = 0.5615  moles

HCl  was  in excess  while  Mg(OH)2  was the  limiting reagent, therefore  moles  of  MgCl2 = 0.180 moles

mass of MgCl2= moles  MgCl2 x molar mass of MgCl2 (  24.3 + 35.5 x2=95.3)

mass =  0.180  x 95.3 = 17.154 grams

Woah that’s a lot to read xd i’ll try though. i’ll get back t you if. i know the andwer

The problem here is that according to aufbau principle before 3d orbital 4s subshell comes.after filling 3p orbital with electrons you have 4 electrons left, you place two in 4s with opposite spin and the other two in 3d in any of its two orbitals of total five orbitals with the same spin.