Answer:
130 Liters
Explanation:
if 1 mol is 22.4 L, then 5.8 mol is 130 L (129.92 but use sig figs)
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Answer:
V KOH = 41 mL
Explanation:
for neutralization:
- ( V×<em>C </em>)acid = ( V×<em>C </em>)base
∴ <em>C </em>H2SO4 = 0.0050 M = 0.0050 mol/L
∴ V H2SO4 = 41 mL = 0.041 L
∴ <em>C</em> KOH = 0.0050 N = 0.0050 eq-g/L
∴ E KOH = 1 eq-g/mol
⇒ <em>C</em> KOH = (0.0050 eq-g/L)×(mol KOH/1 eq-g) = 0.0050 mol/L
⇒ V KOH = ( V×<em>C </em>) acid / <em>C </em>KOH
⇒ V KOH = (0.041 L)(0.0050 mol/L) / (0.0050 mol/L)
⇒ V KOH = 0.041 L
