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Ksenya-84 [330]
3 years ago
11

How many grams of sodium hydroxide are present in 247.0 mL of a 0.300 M NaOH solution?

Chemistry
1 answer:
mamaluj [8]3 years ago
5 0

 The grams  of NaOH  that are  present  in 247.0  ml  of  a 0.300M NaOH solution   is  2.964 grams


<u><em>calculation</em></u>

Step  1: find moles of NaOH

moles  = molarity  x  volume in liters

volume in liters  = 247.0 /1000 =  0.247  liters

molarity =   0.300 M = 0.300  mol/l

moles = 0.300 mol/l   x 0.247 =0.0741  moles



step 2: find mass  of NaOH

mass=  moles x molar mass

The  molar mass of NaOH = 23 +1 +16= 40 g/mol

mass=0.0741  moles x40 g/mol =2.964 grams


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Answer:

9.9 ml of 0.200M NH₄OH(aq)

Explanation:

3NH₄OH(Iaq) + FeCl₃(aq) => NH₄Cl(aq) + Fe(OH)₃(s)

?ml of 0.200M NH₄OH(aq) reacts completely with  12ml of 0.550M FeCl₃(aq)

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1(0.200M)(Vol Am-OH Soln) = 3(0.550M)(0.012L)

=> Vol Am-OH Soln = 3(0.550M)(0.012L)/1(0.200M) = 0.0099 Liter = 9.9 milliliters  

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