Pb(NO3)2+K2CrO4 --> PbCrO4+KNO3 How do you balance this?
1 answer:
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Answer:
1,812 wt%
Explanation:
The reactions for this titration are:
2Ce⁴⁺ + 3I⁻ → 2Ce³⁺ + I₃⁻
I₃⁻ + 2S₂O₃⁻ → 3I⁻ + S₄O₆²⁻
The moles in the end point of S₂O₃⁻ are:
0,01302L×0,03428M Na₂S₂O₃ = 4,463x10⁻⁴ moles of S₂O₃⁻. As 2 moles of S₂O₃⁻ react with 1 mole of I₃⁻, the moles of I₃⁻ are:
4,463x10⁻⁴ moles of S₂O₃⁻× = 2,2315x10⁻⁴ moles of I₃⁻
As 2 moles of Ce⁴⁺ produce 1 mole of I₃⁻, the moles of Ce⁴⁺ are:
2,2315x10⁻⁴ moles of I₃⁻× = 4,463x10⁻⁴ moles of Ce(IV). These moles are:
4,463x10⁻⁴ moles of Ce(IV)× = <em>0,0625 g of Ce(IV)</em>
As the sample has a 3,452g, the weight percent is:
0,0625g of Ce(IV) / 3,452g × 100 = <em>1,812 wt%</em>
I hope it helps!
Answer:
.
Explanation:
Electrons are conserved in a chemical equation.
The superscript of indicates that each of these ions carries a charge of
. That corresponds to the shortage of one electron for each
ion.
Similarly, the superscript on each
ion indicates a shortage of three electrons per such ion.
Assume that the coefficient of (among the reactants) is
, and that the coefficient of
(among the reactants) is
.
.
There would thus be silver (
) atoms and
aluminum (
) atoms on either side of the equation. Hence, the coefficient for
and
would be
and
, respectively.
.
The
ions on the left-hand side of the equation would correspond to the shortage of
electrons. On the other hand, the
ions on the right-hand side of this equation would correspond to the shortage of
electrons.
Just like atoms, electrons are also conserved in a chemical reaction. Therefore, if the left-hand side has a shortage of electrons, the right-hand side should also be
electrons short of being neutral. On the other hand, it is already shown that the right-hand side would have a shortage of
electrons. These two expressions should have the same value. Therefore,
.
The smallest integer and
that could satisfy this relation are
and
. The equation becomes:
.