How many milliliters of 0.025 M H2SO4 are required to neutralize exactly 525 ml of 0.06 M KOH?
Let us first find out how much H2SO4 is required to neutralize 525 ml of 0.06 M KOH.Why ? Because 1 molecule of H2SO4 gives 2 H+ ions per molecule while only one H+ ion is required to neutralize 1 molecule of KOH. So, 1 molecule of H2SO4 can neutralize 2 molecules of KOH. Hence, we would require 525 ml of 0.03 M H2SO4 to neutralize 525 ml of 0.06 M KOH. How will we prepare 525 ml of 0.03 M H2SO4 ?Now, we have 0.025 M H2SO4 and we do not know how much volume we have.We will use the standard N1 X V1 = N2 X V2 for this calculation.N1=0.025 M; V1=unknown; N2=0.03 M and V2=525 ml.So V1= (0.03 X 525)/(0.025) = 630 ml.
How many milliliters of 0.100 M HNO3 are needed to neutralize?
Al(OH)3 + 3 HNO3 ----------> Al(NO3)3 + 3 H2O. 0.0585 L * 0.0100 mol/L = 0.000585 moles Al(OH)3. 3 * 0.000585 moles = 0.001755 moles HNO3 reacted. 1000 mL 0.100 M HNO3 contained 0.100 moles. 0.001755 / 0.100 * 1000 mL = 17.6 mL HNO3 needed.
How many mililiters of 0.100 M HNO3 are needed to neutralize the following solutions?
The balanced equation is: NaOH + HNO3 → NaNO3 + H2O We see that 1 molecule of NaOH reacts with 1 molecule of HNO3, so they react in a one-to-one molar ratio. If we have 34.7 ml. of 0.775 M NaOH solution, that means each liter of NaOH solution contains 0.775 mole of NaOH. We have 0.0347 liter of this solution, so it contains 0.0347 times 0.775, or 0.0268925 mole of NaOH. So we need 0.0268925 mole of HNO3 solution to react with it. If its concentration is 0.1 mole per liter, we will need 0.0268925 mole / 0.1 mole per liter, or 0.268925 liter which is 268.925 ml. (269 ml. to 3 significant digits and the nearest whole milliliter.
How many mL of 0.050 M HNO3 are needed to neutralize 30.0 mL of 0.205 M Ba(OH)2?
Moles Ba(OH)2 = 0.0300 L x 0.205 M = 0.00615 Ba(OH)2 + 2 HNO3 >> Ba(NO3)2 + 2 H2O moles HNO3 needed = 2 x 0.00615 = 0.0123 V = moles / M = 0.0123 / 0.050 = 0.246 L => 246 mL
How many mL of 0.65M HNO3 are needed to neutralize 50 mL of 0.90M Ca(OH)2?
2HNO3 + Ca(OH)2 >> Ca(NO3)2 + 2H2O Moles Ca(OH)2 = 50 x 0.90 /1000 = 0.045 the ratio between HNO3 and Ca(OH)2 is 2 : 1 so moles HNO3 = 0.045 x 2 = 0.090 M = moles / L L = moles/M = 0.090 / 0.65 = 0.138 L = 138 mL
How many mL of .100 M HNO3 are needed to neutralize the following solution?
moles Al(OH)3 = 0.0585 L x 0.100 M = 0.00585 the balanced equation is Al(OH)3 + 3 HNO3 = Al(NO3)3 + 3 H2O moles HNO3 required = 0.00585 x 3 =0.0175 Volume HNO3 = 0.0175 mol/ 0.100 M = 0.175 L = 175 mL
In order to completely neutralize 20 mL of a solution of HCl 0.1 M, 40 mL of a solution of NaOH must be added. What is the M of the NaOH solution?
No, this working is incorrect. You tripped up on two points.20 milliliters is what decimal fraction of a liter?milli- means one thousandth, so 20 mL = 20/1000 liters20/1000 = 2/100 = 1/50 literDivide one by fifty:50 ) 1 wont go , result 0.? so try50) 10 won’t go result 0.0? so try50)100 goes twice result 0.02So 0.02 L = 20mL Result for VaYou made the same mistake with Vb = 40 mL SHOULD BE = 0.04 LThe second mistake was more important because in this equation the first two scaling mistakes canceled out!You wrote: 1 mol x 0.1 M x 0.20 L / 1 mol x 0.40 L = 8x10-3M1 x 0.1 x 0.2 / 1 x 0.4 = 0.05M NOT 8 millimol !You multiplied (x) instead of dividing ( / )
How many mL of 0.20M HNO3 are needed to completely neutralize 35.0mL of 0.10M Ba(OH)2?
Equation: 2HNO3 + Ba(OH)2 → Ba(NO3)2 + 2H2O 2mol HNO3 react with 1 mol BaOH2 Mol Ba(OH)2 in 35.0mL of 0.1M solution = 35/1000*0.1 = 0.0035 mol This will require 0.0035*2 = 0.007 mol HNO3 Volume of 0.2M HNO3 that contains 0.007 mol HNO3 = 0.007/0.2*1000 = 35mL Answer: 35mL of the 0.2M HNO3 solution required.