Answer:
Limiting reactant is NiSO₄
Explanation:
The reaction of aluminum metal with aqueous nickel(II) sulfate to produce aqueous aluminum sulfate and nickel is:
2 Al(s) + 3 NiSO₄ → Al₂(SO₄)₃ + 3 Ni
<em>That means 2 moles of Al react with 3 moles of nickel sulfate.</em>
<em />
Moles of Al and NiSO₄ are:
Al: 108g × (1mol / 26.98g) = 4.00 moles of Al
NiSO₄: 464g × (1mol / 154.75g) = 3.00 moles of NiSO₄
For a complete reaction of aluminium there are necessary:
4.00mol Al ₓ ( 3 moles NiSO₄ / 2 moles Al) = 6 moles of NiSO₄
As you have just 3.00 moles of NiSO₄, the <em>limiting reactant is NiSO₄</em>
The correct equation
for the overall reaction can simply be obtained by adding the two separate
equations together. Now when you add the two equations together, the overall K can
be calculated by multiplying the individual K values. Therefore:<span>
K(overall) = K1 * K2 </span>
K(overall) = (1.6 x
10^-10) * (1.5 x 10^7)
<span>K(overall) = 2.4 x
10^-3</span>
C + O2= CO2
CO2 is limit
5.4-3.72= 1.68 g of C is excess
5.4 g = 100%
3.72 g = x
x=68.9 %
