If 5.0 moles of naoh are mixed with 5.0 moles of h2so4 in a neutralization reaction in aqueous solution, the limiting reactant is 1. na2so4. 2. h2o. 3. naoh. 4. none of the other choices 5. cannot be determined without knowing the volume of the solution 6. h2so4.

the balanced equation for the reaction is

2NaOH + H₂SO₄ ---> Na₂SO₄ + 2H₂O

molar ratio of NaOH to H₂SO₄ is 2:1

limiting reactant is the reagent that is fully consumed in the reaction and determines the amount of product formed and excess reactant is the reactant where only a fraction is used up and theres excess remaining after the reaction

number of NaOH moles added - 5.0 mol

assuming NaOH is the limiting reactant

according to molar ratio, 2 mol of NaOH reacts with 1 mol of H₂SO₄

then 5 mol of NaOH reacts with - 1/2 x 5 = 2.5 mol of H₂SO₄

the number of H₂SO₄ moles added are 5.0 mol

but only 2.5 mol of H₂SO₄ moles is used up therefore H₂SO₄ is the excess reactant and NaOH is limiting reactant and it is fully used up .

answer is 3. NaOH

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"