Answer : The reagent present in excess and remains unreacted is, 
Solution : Given,
Moles of 
= 3.00 mole
Moles of = 2.00 mole
Excess reagent : It is defined as the reactants not completely used up in the reaction.
Limiting reagent : It is defined as the reactants completely used up in the reaction.
Now we have to calculate the limiting and excess reagent.
The balanced chemical reaction is,
From the balanced reaction we conclude that
As, 2 moles of react with 1 mole of
So, 3.00 moles of react with 
moles of
From this we conclude that, is an excess reagent because the given moles are greater than the required moles and is a limiting reagent and it limits the formation of product.
Hence, the reagent present in excess and remains unreacted is,
Answer: 40.1%
Explanation: The mass of calcium in this compound is equal to 40.1 grams because there's one atom of calcium present and calcium has an atomic mass of 40.1 . The molar mass of the compound is 100.1 grams. Using the handy equation above, we get: Mass percent = 40.1 g Ca⁄100.1 g CaCO3 × 100% = 40.1% Ca.
Answer:
I hope this link helps you.
Explanation:
http://astronomy.swin.edu.au/cosmos/P/Phases
Moles = Molarity x Volume
Moles = 2.0 x 0.50
= 1.0 mole
hope this helps!
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.
