Jun 2/26/2018 12:00 AM , 2.1/52/25/2018 10:17 PM Print Calculator-Periodic Table
ID: 711325 • Letter: J
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Jun 2/26/2018 12:00 AM , 2.1/52/25/2018 10:17 PM Print Calculator-Periodic Table uestion 8 of 13 Incorrect Sapling Learning Write the balanced neutralization reaction between H2S0, and KOH in aqueous solution. Phases are optional. H,SO, t. 2KOH K2SO4+2H20 0.550 L of 0.480 M H2SO4 is mixed with 0.500 L of 0.260 M KOH. What concentration of sulfuric acid remains after neutralization? Number 0.278 M H,so, Incorrect If acid is leftover, that means the base is the limiting reactant. The amount of acid leftover is the differes between the amount present and the amount that reacts. Find the number of moles of base present, then use that value and the coefficients from the balanced ch equation to find the number of moles of acid that react. Find the number of moles of acid present, then s the number of moles of acid that react. Finally, use the remaining number of moles of acid and the total to find the final concentration of the acid e Previous Give Up & View Solution # Try Again eNext about us | careers I privacy policy | terms ofuse l contact usExplanation / Answer
The balanced equation is
H2SO4 + 2 KOH -----> K2SO4 + 2 H2O
number of moles of H2SO4 = molarity * volume of solution in L
number of moles of H2SO4 = 0.480 * 0.550 = 0.264 mole
number of moles of KOH = molarity *volume of solution in L
number of moles of KOH = 0.260 * 0.500 = 0.130 mole
from the balanced equation we can say that
1 mole of H2SO4 requires 2 mole of KOH so
0.264 mole of H2SO4 will require 0.528 mole of KOH
but we have only 0.130 mole of KOH so KOH is the limiting reactant
2 mole of KOH requires 1 mole of H2SO4 so
0.130 mole of KOH will require 0.0650 mole of H2SO4
number of moles of H2SO4 left = 0.264 - 0.0650 = 0.199 mol
Therefore, the number of moles of H2SO4 left overe = 0.199 mole
molarity of acid left over = number of moles of acid left over / total volume of solution
molarity of acid left over = 0.199 mol / 1.05 = 0.190 M
Therefore, the molarity of acid left = 0.190 M
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