Answer is: 127 grams <span>rams of metallic copper can be obtained.
</span>Balanced chemical reaction: 2Al + 3CuSO₄ → Al₂(SO₄)₃ + 3Cu.
m(Al) = 54.0 g.
n(Al) = m(Al) ÷ M(Al).
n(Al) = 54 g ÷ 27 g/mol.
n(Al) = 2 mol.
m(CuSO₄) = 319 g.
n(CuSO₄) = 319 g ÷ 159.6 g/mol.
n(CuSO₄) = 2 mol; limiting reactant.
From chemical reaction: n(CuSO₄) : n(Cu) = 3 : 3 (1 : 1).
n(Cu) = 2 mol.
n(Cu) = 2 mol · 63.55 g/mol.
n(Cu) = 127.1 g.
Answer:
The charge carried by each ion (oxidation state of each atom)
Explanation:
If we have an ionic compound and we want to write its formula, we must first know the magnitude of charge on each ion (shown as oxidation state of the atoms involved) because the magnitude of charge on each ion is eventually crisscrossed and gives the subscript (number of atoms) for each atom in the formula.
For instance, let us write the formula of calcium bromide. Ca has a charge of +2 while Br has a charge of -1. If we exchange the charges and ignore the signs such that the crisscrossed charges form subscripts we can now write;
.
Mass of Li = 237.38 g
<h3>Further explanation</h3>
The mole itself is the number of particles contained in a substance amounting to 6.02.10²³

<h3>Known</h3>
Moles of Li = 34.2
Molar mass(MW) of Li = 6.941 g/mol
then mass of Lithium (Li) :

Window cleaner is ammonia dissolved in water. This is an example of a _______.
Answer:
Explanation:
There are some mistakes in writing the equation. However, Please check the below one and contact me again if necessary.
2KI(aq) + Pb(NO₃)₂ (aq) -> PbI₂ (s) + 2KNO₃