Show work and inlcude units on each number sulfuric acid reacts with sodium hydroxide. if 3.50 mol h2so4 is made to react with 10.0 mol naoh, identify the limiting reagent. first write the balanced equation, then do the calculations. show all work!

 The  limiting  reagent  is   H₂SO₄

   calculation

Step 1 :write the equation for reaction

2 NaOH + H₂SO₄  → Na₂SO₄   + 2 H₂O

Step  2: use the mole ratio to determine the  moles of product  produced from each reactant

that is   from  equation above,

NaOH : Na₂SO₄  is 2 :1 therefore the  moles of Na₂So₄

= 10.0 moles x 1/2 =  5.0 moles

H₂SO₄ :Na₂SO₄  is 1:1 therefore the moles  of Na₂SO₄  is also = 3.50 moles

H₂SO₄  is the limiting reagent since  it produces  less amount of Na₂SO₄

atomic mass = 39

number of neutrons = 20

number of protons= 19

electronic configuration = 1s2 2s2 2p6 3s2 3p6 4s1

the number of electron in first shell = 2

number of electrons in n=3 shell = 8