# of atoms per mol = Avogadro’s # (6.022 x 10^23)
Number of mols = mass of substance / molar mass
73 g / 40.08 g = 1.8 mols of Ca in 73 grams
1.8 mols x avagadro’s # = 1.1 x 10^24 atoms in 73 grams of Ca
It is a reaction between an element and a compound that results in a different element and compound
2..............................................
when heat gained = heat lost
when AL is lost heat and water gain heat
∴ (M*C*ΔT)AL = (M*C*ΔT) water
when M(Al) is the mass of Al= 225g
C(Al) is the specific heat of Al = 0.9
ΔT(Al) = (125.5 - Tf)
and Mw is mass of water = 500g
Cw is the specific heat of water = 4.81
ΔT = (Tf - 22.5)
so by substitution:
∴225* 0.9 * ( 125.5 - Tf) = 500 * 4.81 * (Tf-22.5)
∴Tf = 30.5 °C
Ans: 15.1 grams
Given reaction:
Na2CO3 + Ca(OH)2 → 2NaOH + CaCO3
Mass of Na2CO3 = 20.0 g
Molar mass of Na2CO3 = 105.985 g/mol
# moles of Na2CO3 = 20/105.985 = 0.1887 moles
Based on the reaction stoichiometry: 1 mole of Na2CO3 produces 2 moles of NaOH
# moles of NaOH produced = 0.1887*2 = 0.3774 moles
Molar mass of NaOH = 22.989 + 15.999 + 1.008 = 39.996 g/mol
Mass of NaOH produced = 0.3774*39.996 = 15.09 grams
