If 0.340 mol of solid CaCO3 and 410 mL of 0.829 M aqueous H2SO4 are reacted stoichiometrically according to the balanced equation, how many milliliters of 0.829 M aqueous H2SO4 remain? Round your answer to 3 significant figures.
H2SO4(aq) + CaCO3(s) → CO2(g) + CaSO4(s) + H2O(l)
6. Determine Keq for the reaction: 2 SO2 (g) + O2(g) 2 SO3(g), given that 1.00 x 10-2 moles of SO2 and 2.00 x 10-2 moles of O2 were placed in a 2.00L reaction chamber. The chamber contained 7.5 x 10-3 moles of SO3 when equilibrium was established at 727oC.
(PV = nRT, R = 0.0821L• atm/mol • K)
To find out how many milliliters of 0.829 M aqueous H2SO4 remain after the reaction, we first need to determine the limiting reactant.
To do this, we need to compare the number of moles of H2SO4 and CaCO3. The balanced equation tells us that 1 mole of CaCO3 reacts with 1 mole of H2SO4. Therefore, the number of moles of H2SO4 is equal to the number of moles of CaCO3.
Given that we have 0.340 mol of CaCO3, we can determine the number of moles of H2SO4:
0.829 M is the same as 0.829 mol/L (mole per liter). Therefore, the number of moles of H2SO4 is:
0.829 mol/L * 0.410 L = 0.339 mol
Since the number of moles of H2SO4 (0.339 mol) is slightly less than the number of moles of CaCO3 (0.340 mol), we can conclude that H2SO4 is the limiting reactant. This means that all of the H2SO4 will be used up in the reaction, and some of it will remain.
Now, we need to determine the number of moles of H2SO4 that reacted with CaCO3. Since the stoichiometry is 1:1, the number of moles of H2SO4 that reacted is also 0.340 mol.
Next, we need to determine the volume of the 0.829 M aqueous H2SO4 solution that contains 0.340 mol of H2SO4. We can use the equation:
Molarity (M) = moles (mol) / volume (L)
Rearranging the equation, we can solve for the volume:
volume (L) = moles (mol) / Molarity (M)
Therefore, the volume of the 0.829 M aqueous H2SO4 solution that contains 0.340 mol of H2SO4 is:
volume = 0.340 mol / 0.829 M ≈ 0.410 L or 410 mL (rounded to 3 significant figures)
Since no more H2SO4 is available after the reaction, the amount of remaining H2SO4 is zero. Hence, after the reaction, there will be no milliliters of 0.829 M aqueous H2SO4 remaining.
So, the answer is 0 mL.