a. AlCl₃ ⇒ limiting reactant(smaller ratio)
Cu ⇒ excess reactant
b. the mass of leftover reactant : 7.207 g
<h3>Further explanation</h3>
Given
25 g Cu
25 g AlCl3
Required
a. the excess and limiting reactants
b. the mass of leftover reactant
Solution
Reaction
3Cu + 2AlCl₃ ⇒ 3CuCl₂ + 2Al
mol Cu(Ar = 63.5 g/mol) :
mol = mass : Mw
mol = 25 : 63.5
mol = 0.394
mol AlCl3(MW=133,34 g/mol) :
mol = 25 : 133,34 g/mol
mol = 0.187
mol ratio to reaction coefficient Cu : AlCl₃ =
![\tt \dfrac{0.394}{3}\div \dfrac{0.187}{2}=0.131\div 0.093](https://tex.z-dn.net/?f=%5Ctt%20%5Cdfrac%7B0.394%7D%7B3%7D%5Cdiv%20%5Cdfrac%7B0.187%7D%7B2%7D%3D0.131%5Cdiv%200.093)
AlCl₃ ⇒ limiting reactant(smaller ratio)
Cu ⇒ excess reactant
b. the mass of leftover reactant :
mol Cu = 3/2 x 0.187 = 0.2805
mol left = 0.394 - 0.2805 = 0.1135
mass = 0.1135 x 63.5 = 7.207 g
7 different types of tide
<span>
"1 x Ca = 1 x 40 = 40
1 x C = 1 x 12 = 12
3 x O = 3 x 16 = 48
40 + 12 + 48 = 100 (Mr of CaCO3)
moles = mass / Mr
mass = moles x Mr
mass = 1 mole x 100 = 100grams."</span><span>
</span>
Silicon and Germanium. You can conclude it from its position in the periodic table. Silicon and Germanium are in the same group of the Carbon. That implies that all three have the same number of electrons in the outer orbital, which is responsible for the bonds.