Molar mass of KOH = 39.1+16.0+1.0 = 56.1
Amount of KOH = mass of sample / molar mass = 32/56.1 = 0.570mol
1 mole KOH = 56.105g KOH
234.1g KOH x 1mol KOH/56.105g KOH = 4.171 moles KOH
Molarity = moles of solute/Liters of solution ( 220.0 ml = 0.220 Liters ) 0.500 M KOH = moles KOH/0.220 Liters = 0.110 moles KOH (56.108 grams/1 mole KOH) = 6.17 grams solid KOH needed
Molarity = moles of solute/Liters of solution get moles KOH 6.31 grams KOH (1 mole KOH/56.108 grams) = 0.11246 moles KOH 0.250 M KOH = 0.11246 moles KOH/XL 0.11246/0.250 = 0.4498 liters = 450 milliliters
I assume you mean 7.25 M KOH, but the process is the same anyway.Molarity = moles of solute/Liters of solution ( 400 ml = 0.4 Liters )7.25 M KOH = X moles KOH/0.4 Liters= 2.9 moles KOH (56.108 grams/1 mole KOH)= 163 grams potassium hydroxide needed===============================
Depending upon how precise your masses are, KOH has a molar mass of about 56, so your answer is about 34/56.
I assume KOH is limiting. Balanced equation. KOH + HCl -> KCl + H2O 0.400 moles KOH (1 mole H2O/1 mole KOH)(18.016 grams/1 mole H2O) = 7.21 grams water produced =====================
Molarity = moles of solute/Liters of solution ( 220.0 ml = 0.220 Liters ) 0.500 M KOH = moles KOH/0.220 Liters = 0.110 moles KOH (56.108 grams/1 mole KOH) = 6.17 grams solid KOH needed
Molarity = moles of solute/Liters of solution get moles KOH 6.31 grams KOH (1 mole KOH/56.108 grams) = 0.11246 moles KOH 0.250 M KOH = 0.11246 moles KOH/XL 0.11246/0.250 = 0.4498 liters = 450 milliliters
I assume you mean 7.25 M KOH, but the process is the same anyway.Molarity = moles of solute/Liters of solution ( 400 ml = 0.4 Liters )7.25 M KOH = X moles KOH/0.4 Liters= 2.9 moles KOH (56.108 grams/1 mole KOH)= 163 grams potassium hydroxide needed===============================
Depending upon how precise your masses are, KOH has a molar mass of about 56, so your answer is about 34/56.
Moles KOH = Molarity x Volume = 0.214 moles/liter x 0.0602 liters = 0.0129 moles KOH. Remember, 60.2 mL = 0.062L
I assume KOH is limiting. Balanced equation. KOH + HCl -> KCl + H2O 0.400 moles KOH (1 mole H2O/1 mole KOH)(18.016 grams/1 mole H2O) = 7.21 grams water produced =====================
1 mole of KOH is 56.1074 grams1.350 g/56.1074 g =0.02406 (2.406 x 10^-2) moles of KOH
49.7g O2
Molarity = moles of solute/liters of solution ( 300 ml = 0.300 liter ) 0.250 molar KOH = moles KOH/0.300 liters = 0.075 moles KOH
Two moles KOH for one mole Mg(OH)2; so for 4 moles KOH - two moles Mg(OH)2.And two moles Mg(OH)2 is equal to 116,64 g.
Two moles KOH for one mole Mg(OH)2; so for 4 moles KOH - two moles Mg(OH)2.And two moles Mg(OH)2 is equal to 116,64 g.
Molarity= Number of moles of solute/Liters of solution 50 grams KOH 700 ML to .7 Liters of h2o Molar Mass of KOH= 56 50 divided by 56 = .89 moles Molarity= .89 mol/.7 L = 1.27 MOLARITY