The delta H of -484 kJ is the heat given off when 2 moles of H2 react with 1 mole of O2 to make 2 moles of H2O. You don't have anywhere near that much reactants, only 1/4 as much
<span>actual delta H = 0.34 moles H2 x (-484 kJ / 2 moles H2) = 823 kJ </span>
<span>delta E = delta H - PdeltaV = 823 kJ - 0.41 kJ = 822 kJ</span>
Answer:
Molarity of Sr(OH)₂ = 0.47 M
Explanation:
Given data:
Volume of Sr(OH)₂ = 15.0 mL
Volume of HCl = 38.5 mL (0.0385 L)
Molarity of HCl = 0.350 M
Concentration/Molarity of Sr(OH)₂ = ?
Solution:
Chemical equation:
Sr(OH)₂ + 2HCl → SrCl₂ +2H₂O
Number of moles of HCl:
Molarity = number of moles/ volume in L
0.350 M = number of moles/0.0385 L
Number of moles = 0.350 mol/L× 0.0385 L
Number of moles = 0.0135 mol
Now we will compare the moles of HCl with Sr(OH)₂.
HCl : Sr(OH)₂
2 : 1
0.0135 : 1/2×0.0135 = 0.007 mol
Molarity/concentration of Sr(OH)₂:
Molarity = number of moles / volume in L
Molarity = 0.007 mol /0.015 L
Molarity = 0.47 M
The balanced chemical reaction is expressed as:
<span>8SO2 + 16H2S = 3S8 + 16H2O
We are given the initial amount of the reactants. From there, we determine the limiting reactant. We do as follows:
87.0 g SO2 ( 1 mol / 64.07 g ) = 1.36 mol SO2 ( 16 mol H2S / 8 mol SO2 ) = 2.72 mol H2S
87.0 g H2S ( 1 mol / 34.08 g ) = 2.55 mol H2S ( 8 mol SO2 / 16 mol H2S ) = 1.28 mol SO2
Therefore, the limiting reactant would be H2S. We calculate the maximum amount of S8 that can be produced from the amount of H2S.
2.55 mol H2S ( 3 mol S8 / 16 mol H2S ) ( 256.48 g / 1 mol ) = 122.63 g S8</span>
Everything dead would just keep piling up and up and up and nothing would ever go away! The Earth would be like a giant landfill! <span />
