Question

HCl + NaOH → NaCl + H2O

Answer 1

Answer 1: 
The reaction involved in present case is
                                           HCl + NaOH → NaCl + H2O
In present case, it can be seen that 1 mole of HCl reacts with 1 mole of NaOH to give 1 mole of NaCl and H2O.

We know that, number of mole = weight of substance (g)/molecular weight

For HCl, number of mole = 30/36.5 = 0.8219
For NaOH, number of mole that can be produced = 0.8219

Thus, theoretical yield of NaOH = number of mole X Molecular weigh       
                                                   = 0.8219 X 58.5
                                                   = 48.08 g.

Theoretical yield of the experiment is 48.08 g.
.......................................................................................................................
Answer 2:
Percent yield is mathematically expressed as,
$\frac{\text{Actual Yeild}}{\text{Theoretical Yield}} X 100$

In present case, Actual Yield = 10 g
Theoretical Yield = 48.08 g.

Therefore,  Percent yield = $\frac{\text{10}}{\text{48.08}} X 100$
                                        = 20.79%

Thus, percent yield of the experiment is 20.79 %

Answer 2

HCl + NaOH  = NaCl  +H2O
          Question  1
 if  30g of HCl  is reacted  with excess  NaOH  and 10g of  NaCl  is  produced.The theoretical  yield  of experiment  is  48.08  g

    calculation
find the  moles  of  HCl  used=  mass / molar mas

= 30 g/36.5 = 0.8219   moles

by  use of mole ratio  between  HCl  to NaCl  which is 1:1  the moles  of  NaCl  is  = 0.8219 moles

therefore the  theoretical  mass  of NaCl  =  moles x  molar   mass  of NaCl

=  0.8219  moles  x  58.5  g/mol =  48.08  grams

                    
                      Question 2

The % mass  =  20.8%

            calculation

% mass =  actual yield /  theoretical  yield  x 100

actual yield = 10 g
theoretical  yield = 48.08 g

%  yield is therefore  = 10 g/  48.08 g  x100 =  20.8%