A 1.00g sample of a metal X (that is known to form X3+ ions in solution) was add
ID: 576087 • Letter: A
Question
A 1.00g sample of a metal X (that is known to form X3+ ions in solution) was added to 127.9ml of 0.5000M sulfuric acid. After all the metal had reacted, the remaining acid required 0.03340L of 0.5000M NaOH solution for complete neutralization. Calculate the molar mass of the metal and identify the element. A 1.00g sample of a metal X (that is known to form X3+ ions in solution) was added to 127.9ml of 0.5000M sulfuric acid. After all the metal had reacted, the remaining acid required 0.03340L of 0.5000M NaOH solution for complete neutralization. Calculate the molar mass of the metal and identify the element.Explanation / Answer
2NaOH + H2So4 -----------------> Na2SO4 + 2H2O
no of moles of NaOH = molarity * volume in L
= 0.5*0.03340 = 0.0167 moles
2 moles of NaOH react with 1 mole of H2So4
0.0167 moles of NaOH react with = 1*0.0167/2 = 0.00835 moles of H2SO4 (excess no of moles of H2So4)
no of moles of H2SO4 = molarity*volume in L
= 0.5*0.1279 = 0.06395 moles
reacted no of moles of H2So4 = 0.06395-0.00835 = 0.0556moles of H2So4 reacted with metal
2X + 3H2So4 ------------------> X2(SO4)3 + 3H2
3 moles of H2So4 react with 2 moles of X
0.0556 moles of H2So4 react with = 2*0.0556/3 = 0.037 moles of X
mass of metal 1g
molar mass of metal X = mass of metal X/no of moles of X
= 1/0.037 = 27g/mole
That metal is aluminum
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