How do you do Stoichiometry

We are doing mole conversion and stoichiometry in class, I understand how to do mole conversion but I dont understand the stoichiometry part of the homework. This is one of the questions.

2NaOH+H2SO -> 2H2O+Na2SO4
How many grams of H2O should be formed if you start with 75g of NaOH?

Suppose the above reaction produced 28g of H2O, What is the percent Yield?

Can someone help show me how to do this step by step

So stoichometry basically states that we know 2 moles of NaOH will give us 2 moles of H20 in this case.

so,

75 g NaOH x 1 mole NaOH/40 g x 2 moles H20/ 2 moles NaOH x 18 grams H20/1 mole H20 = 33.75 grams of H20 is our theoretical yield. Now, we know that the percent yield is the actual over theoretical.

(28/33.75) x 100 for a percent = 83%

1. 75.0-g NaOH x( 1 mol NaOH / 40-g NaOH) x( 2 mol H20 / 2 mol NaOH) x (18.0-g H2O / 1 mol H2O)= 33.8-g H20

2. % Yield = 28-g/33.8-g x 100% = 82.8 %

3 stoichiometry: a. divide mass of given material by molar mass to get moles b.) use molar ratio of starting and ending material (NaOH ---> H2O) from coefficients in balanced equation (2:2) c.) multiply by molar mass of ending material

% yield - divide actual yield by theoretical yield then multiply by 100 to convert to a perccent.

2NaOH+H2SO4 --> 2H2O+Na2SO4

If we assume the H2SO4 is in excess the moles of H2O are limited by the mass of NaOH

Molar mass of NaOH = 40g so moles of NaOH = 75 / 40 = 1.875 moles

we notice 2 moles of NaOH gives 2 moles of water so moles of water produced by
1.875 moles of NaOH = 1.875 moles

molar mass of H2O is 18 g / mole so mass of water = 1.875 x 18 = 33.75 g

if we have 28 g of water the % yield = 28g/ 33.75 g = 83.0%

So, to start off you want to convert grams of NaOH into moles of NaOH using the compounds molar mass.