T Thom New member May 3, 2009 #1 Im struggling with this: 25cm^3 of H2SO4 are placed in a conical flask and neutrilised with 8.70cm^3 of NaOH at a consentration of 0.150 mol dm^-3. H2SO4 +2NaOH --> Na2SO4 +2H2O Calculate the amount in moles of the Sodium Hydroxide used!?
Im struggling with this: 25cm^3 of H2SO4 are placed in a conical flask and neutrilised with 8.70cm^3 of NaOH at a consentration of 0.150 mol dm^-3. H2SO4 +2NaOH --> Na2SO4 +2H2O Calculate the amount in moles of the Sodium Hydroxide used!?
D Dr.A New member May 3, 2009 #2 moles NaOH = 0.00870 dm^3 x 0.150 mol dm^-3 =0.00131 Moles H2SO4 = 0.00131/2 =0.000655 molarity H2SO4 = 0.000655 mol/ 0.025 dm^3 =0.0262 M
moles NaOH = 0.00870 dm^3 x 0.150 mol dm^-3 =0.00131 Moles H2SO4 = 0.00131/2 =0.000655 molarity H2SO4 = 0.000655 mol/ 0.025 dm^3 =0.0262 M
