calculate number of moles of h2so4 needed to react with 25.0ml of .20M liOH
Don't get so lazy in chemistry you can't use the caps key. CO, Co, and co all mean something different in chemistry just as m and M are not the same.
H2SO4 + LiOH ==> Li2SO4 + 2H2O
mols LiOH = M x L = ?
Use the coefficients in the balanced equation to covert mols LiOH to mols H2SO4.
mols H2SO4 = ?
0.0025
To calculate the number of moles of H2SO4 needed to react with 25.0 mL of 0.20 M LiOH, we can use the concept of stoichiometry. The balanced chemical equation for the reaction between H2SO4 and LiOH is:
H2SO4 + 2LiOH → Li2SO4 + 2H2O
The equation shows that 1 mole of H2SO4 reacts with 2 moles of LiOH.
First, we need to calculate the number of moles of LiOH present in the given volume (25.0 mL) and concentration (0.20 M):
Moles of LiOH = Volume (in liters) × Concentration
Volume = 25.0 mL = 25.0 mL × (1 L / 1000 mL) = 0.0250 L
Concentration = 0.20 M
Moles of LiOH = 0.0250 L × 0.20 mol/L = 0.00500 moles
According to the balanced equation, 1 mole of H2SO4 reacts with 2 moles of LiOH. Therefore, for every 2 moles of LiOH, we need 1 mole of H2SO4.
Hence, the number of moles of H2SO4 needed = 0.00500 moles ÷ 2 = 0.00250 moles
Therefore, 0.00250 moles of H2SO4 are needed to react with 25.0 mL of 0.20 M LiOH.