How many milliliters of 0.205 M NH3 solution will exactly react with 50.0 mL of 0.150 M H2SO4 solution?
2NH3(aq) + H2SO4(aq) → (NH4)2SO4(aq)
moles H2SO4 = M x L = ??
Using the equation coefficients, convert moles H2SO4 to moles NH3.
M NH3 = moles NH3/L NH3.
You know M and moles, solve for L and convert to mL.
To find the number of milliliters of 0.205 M NH3 solution that will react with 50.0 mL of 0.150 M H2SO4 solution, you first need to determine the stoichiometry of the balanced chemical equation.
The balanced chemical equation is:
2NH3(aq) + H2SO4(aq) → (NH4)2SO4(aq)
From the equation, you can see that it takes two moles of NH3 to react with one mole of H2SO4.
Next, calculate the number of moles of H2SO4 using its concentration and volume given:
Moles of H2SO4 = concentration of H2SO4 × volume of H2SO4 solution
= 0.150 M × 0.0500 L (convert mL to L)
= 0.00750 moles
Since the stoichiometry of the balanced chemical equation is 2:1 between NH3 and H2SO4, you need half the number of moles of NH3 to react. Therefore:
Moles of NH3 = 0.00750 moles ÷ 2
= 0.00375 moles
Finally, calculate the volume of 0.205 M NH3 solution using its molarity and the number of moles calculated:
Volume of NH3 solution = moles of NH3 ÷ concentration of NH3
= 0.00375 moles ÷ 0.205 M
≈ 0.0183 L (convert to mL)
≈ 18.3 mL
Therefore, approximately 18.3 milliliters of the 0.205 M NH3 solution will react exactly with 50.0 mL of the 0.150 M H2SO4 solution.