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
Answer:
No. Of Moles of zinc = m/Ar
= 13/ 65.38 = 0.198 moles
From balanced equation, Mole ration between CuSO4 and Zn is 1 : 1
So only 0.198 moles of CuSO4 reacts, it is in excess
Mass = no of Moles X Mr
Mass = 0.198 X 159.5 = 31.59 grams
Volume = mass m denisty
Volume j 31.59 / 3.6 = 8.78 ml
Explanation:
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