50.0mL of 1.00M Hcl and 2.0 g of NaOH. What is limiting reagant....?

Question

50.0mL of 1.00M Hcl and 2.0 g of NaOH. What is limiting reagant.... how many grams of water will be produced and how many moles of the excess reagant will be left? Please show work.

Answer 1

HCl+NaOH >>NaCl+H2O

2 g/ 40=0.1 mole OH-

moles H+ = 0.05 limiting reactant

we get 0.05 moles H2O >> 0.9 g

Answer 2

1L of HCL 1M contains 1 mole
so 50mL = 0.05 L = 0.05mole

2g of NaOH MW NaOH =23+1+17= 40g
and 2g of NaOH = 2/40=0.05mole

These two components are in stoechiometric agreement

NaOH + HCl = NaCl + H2O

you will produce 0.05mole of H2O = 0.05 *18 = 0.9g

No excess will be left