what is the excess reactant of H2SO4+2NaOH->Na2SO4+2H2O

H2SO4: NaOH:
n=? n=?
c=0.025M c=0.030M
v=0.4L v=0.4

This site doesn't handle spaces very well. I assume you mean 0.4 L of 0.025M H2SO4 with 0.4 L of 0.030M NaOH.

mols H2SO4 = M x L = approx 0.01
mols NaOH = M x L = approx 0.012

For 0.01 mols H2SO4 you need 0.02 mols NaOH and you don't have that much so NaOH must be the limiting regent. That makes H2SO4 the excess reagent.

To find the excess reactant in the given chemical equation H2SO4 + 2NaOH -> Na2SO4 + 2H2O, we can follow these steps:

Step 1: Calculate the number of moles of each reactant using the given formula:

n = c x v

For H2SO4:
n = 0.025 M x 0.4 L = 0.01 moles

For NaOH:
n = 0.030 M x 0.4 L = 0.012 moles

Step 2: Determine the stoichiometric ratio between the reactants.

From the balanced chemical equation, we can see that the ratio between H2SO4 and NaOH is 1:2. This means that 1 mole of H2SO4 reacts with 2 moles of NaOH.

Step 3: Calculate the moles required for the reaction based on the stoichiometric ratio.

Since the stoichiometric ratio is 1:2, we need twice as many moles of NaOH as H2SO4.

Moles of NaOH required = 2 x moles of H2SO4 = 2 x 0.01 moles = 0.02 moles

Step 4: Compare the moles of each reactant with the moles required for the reaction.

For H2SO4:
Moles available = 0.01 moles
Moles required = 0.02 moles

For NaOH:
Moles available = 0.012 moles
Moles required = 0.02 moles

Step 5: Determine the excess reactant.

The reactant with the higher number of moles available is the excess reactant.

In this case, NaOH has more moles available (0.012 moles) compared to the moles required (0.02 moles). Therefore, NaOH is the excess reactant.

To determine the excess reactant in a chemical reaction, we need to compare the stoichiometry of the reaction with the amounts of reactants given.

First, let's determine the number of moles (n) for each reactant using the formula:

n = c * v

where c is the concentration and v is the volume.

For H2SO4:
n(H2SO4) = 0.025 M * 0.4 L
n(H2SO4) = 0.01 moles

For NaOH:
n(NaOH) = 0.030 M * 0.4 L
n(NaOH) = 0.012 moles

Now we can compare the stoichiometry of the reaction with the amounts of reactants.

From the balanced equation:
H2SO4 + 2NaOH -> Na2SO4 + 2H2O

The stoichiometry tells us that 1 mole of H2SO4 reacts with 2 moles of NaOH. Therefore, a 1:2 ratio.

If we compare the amounts of reactants given, we can see that there is less H2SO4 (0.01 moles) than NaOH (0.012 moles). This means that H2SO4 is the limiting reactant.

The excess reactant is the one that does not get fully consumed in the reaction, which in this case is NaOH.

Thus, the excess reactant is NaOH.