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Question 1 of 9

A solution contains 5.46 grams or 35 millimoles of Sodium Acid Phosphate BP. What is the RMM to one decimal place of Sodium Acid Phosphate BP?
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grams
The RMM of Sodium Acid Phosphate BP is 156.0.

Convert millimoles to moles: 35 mmoles = 0.001 x 35 mmoles = 0.035 moles

RMM =     Mass of Component    
Number of Moles
RMM of Sodium Acid Phosphate BP =     5.46    
0.035

RMM of Sodium Acid Phosphate BP = 156.00

RMM of Sodium Acid Phosphate = 156.0 to one decimal place

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Question 2 of 9

The following reaction takes place:

NaOH + HCl ----» NaCl + H2O

If 4.0 grams of Sodium Hydroxide were used and there was more than enough Hydrochloric Acid present, how many grams of Sodium Chloride were produced?

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grams
5.85 grams of Sodium Chloride were produced.
RMM NaOH = 23.0 + 16.0 + 1.0 = 40.0            RMM NaCl = 23.0 + 35.5 = 58.5
  1. Find out how many moles of NaOH were used:
    Moles =    Mass of Component   
    RMM
    Moles of Sodium Hydroxide = 4.0 / 40.0 = 0.1 moles
  2. Find out how many moles of NaCl are produced for every one mole of NaOH used.
    From the equation, you can see that one mole of NaOH will result in one mole of NaCl.
  3. Find the number of moles of NaCl produced.
    Moles of NaCl = 1 x Number of NaOH moles used
    Moles of NaCl = 0.1 moles
    Therefore mass of NaCl produced is Moles x RMM
    Mass of NaCl produced = 0.1 x 58.5 grams
    Mass of NaCl produced = 5.85 grams
  1. Find out how many moles of Sodium Hydroxide were used
  2. Find out how many moles of Sodium Chloride are produced for every one mole of Sodium Hydroxide used
  3. Find the number of moles of Sodium Hydroxide produced

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Question 3 of 9

A solution containing 38.6 grams of Magnesium Hydroxide (RMM 58.3) is mixed with a solution containing 21.9 grams of Hydrochloric Acid (RMM 36.5) and the following reaction occurs:
Mg(OH)2 + 2 HCl ----» MgCl2 + 2 H2O
How many grams of Magnesium Chloride (RMM 95.3) are produced from the reaction?
Type in your answer and PRESS ENTER.
grams
28.59 grams of Magnesium Chloride are produced.
1 & 2 Find out which of the reagents will limit the reaction and how many moles of it there are.
Number of Moles =  Mass of Component 
RMM
Moles of Magnesium Hydroxide = 38.6/58.3 = 0.66 moles
Moles of Hydrochloric Acid = 21.9 / 36.5 = 0.6 moles
For every 1 mole of Magnesium Hydroxide used 2 moles of Hydrochloric Acid are used.
Hydrochloric Acid is the limiting factor and there are 0.6 moles.

3 & 4 Find out how many moles of Magnesium Chloride are produced for every mole of Hydrochloric Acid used. From the equation, you can see:
For every 2 moles of HCl used, 1 mole of Magnesium Chloride is produced. Thus:
For every 1 mole of HCl used, 0.5 moles of Magnesium Chloride are produced
Therefore Moles of Magnesium Chloride produced = 0.5 x 0.6 = 0.3 moles
Therefore Mass of Magnesium Chloride = 0.3 x 95.3 g = 28.59 grams

  1. Find out which of the reagents will limit the reaction
  2. Find out how many moles of the limiting reagent are used
  3. Find out how many moles of Magnesium Chloride are produced for every mole of limiting reagent used
  4. Find the number of moles of Magnesium Chloride produced.

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Question 4 of 9

A solution containing 23.6 grams of Succinic Acid and a solution containing 35.0 grams of Sodium Hydroxide are mixed and the following reaction occurs:

(CH2COOH)2 + 2 NaOH -----» (CH2COONa)2 + 2 H2O
How many grams of Sodium Succinate are produced?
Type in your answer and PRESS ENTER.
grams
32.4 grams of Sodium Succinate are produced.

RMM Succinic Acid = 118.0    RMM Sodium Hydroxide = 40.0    RMM Sodium Succinate = 162.0

1 & 2 Find out which of the reagents will limit the reaction and how many moles of it there are.

Number of Moles =  Mass of Component 
RMM

Moles of Succinic Acid     = 23.6 / 118.0 = 0.2 moles
Moles of Sodium Hydroxide    = 35.0 / 40.0 = 0.875 moles
Succinic Acid is the limiting reagent and there are 0.2 moles.

3 & 4 Find out how many moles of Sodium Succinate are produced for every mole of Succinic Acid used.
From the equation, you can see:

For every 1 mole of Succinic Acid used, 1 mole of Sodium Succinate is produced.

Therefore Moles of Sodium Succinate produced = 0.2 moles

Therefore Mass of Sodium Succinate = 0.2 x 162.0 g = 32.4 grams

  1. Find out which of the reagents will limit the reaction
  2. Find out how many moles of the limiting reagent are used
  3. Find out how many moles of Magnesium Chloride are produced for every mole of limiting reagent used
  4. Find the number of moles of Magnesium Chloride produced.

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Question 5 of 9

If there are 39.5 grams of Potassium Permanganate (KMnO4) in 100 ml of solution, how many moles of Potassium Permanganate are there in 1000 ml of solution?

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ml
There are 2.5 moles of Potassium Permanganate in 1000 ml.
RMM KMnO4 = 158.0 EITHER:
  1. Find out how many moles there are in 100 ml
    Moles of Potassium Permanganate = 39.5 / 158.0 = 0.25 moles
  2. Then, using ratios find how many moles there are in 1000 ml
    100 ml of solution contains 0.25 moles of Potassium Permanganate
    1000 ml of solution contains  1000  x 0.25 moles = 2.5 moles
    100
OR:
  1. Using ratios, find how many grams there are in 1000 ml
    100 ml of solution contains 39.5 grams of Potassium Permanganate
    1000 ml of solution contains  1000  x 39.5 = 395 grams of Potassium Permanganate
    100
  2. Then find how many moles there are in 1000 ml
    Moles of Potassium Permanganate = 395 / 158.0 moles = 2.5 moles
EITHER:
  1. Find out how many moles there are in 100 ml
  2. Then, using ratios, find how many moles there are in 1000 ml
OR
  1. Using ratios, find how many grams there are in 1000 ml
  2. Then find how many moles there are in 1000 ml

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Question 6 of 9

If there are 0.75 moles of Ammonia (NH3) in 1000 ml of solution, how many grams - to two decimal places - of Ammonia do you need if preparing 250 ml of the solution?
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grams
You need 3.19 grams of Ammonia for 250 ml of solution.
RMM NH3 = 17.0 EITHER:
  1. Find how many grams of Ammonia there are in 1000 ml
    Mass of Ammonia in 1000 ml = 0.75 x 17.0 = 12.75 grams
  2. Using ratios, find how many grams of Ammonia there are in 250 ml
    1000 ml of solution contains 12.75 grams of Ammonia
    250 ml of solution contains   250    x 12.75 grams = 3.187 g = 3.19 g to 2 decimal places
    1000
OR:
  1. Using ratios, find how many moles of Ammonia there are in 250 ml of solution
    1000 ml of solution contains 0.75 moles of Ammonia
    250 ml of solution contains 250 / 1000 x 0.75 moles = 0.1875 moles of Ammonia
  2. Find how many grams of Ammonia there are in 250 ml
    Mass of Ammonia = 0.1875 x 17.0 g = 3.187 g = 3.19 g to 2 decimal places
EITHER:
  1. Find out how many grams of Ammonia there are in 1000 ml
  2. Using ratios, find how many grams there are in 250 ml
OR
  1. Using ratios, find how many moles of Ammonia there are in 250 ml
  2. Find how many grams of Ammonia there are in 250 ml

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Question 7 of 9

What is the RMM of Calcium Chloride Dihydrate if 0.3 moles weigh 44.13 grams? Answer to 1 decimal place.
Type in your answer and PRESS ENTER.
The RMM of Calcium Chloride Dihydrate is 147.1.
RMM =   Mass of Component  
Number of Moles

RMM of Calcium Chloride Dihydrate = 44.13
0.3

RMM of Calcium Chloride Dihydrate = 147.1

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Question 8 of 9

If 1.95 moles of Sucrose (RMM 342.0) are associated with 333 grams of water, how many grams - to one decimal place - of Sucrose are associated with 1000 grams of water?
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grams
2002.7 g of Sucrose are associated with 1000 g of water.
EITHER:
  1. Find how many grams of Sucrose are associated with 333 g water
    Mass of Sucrose = 1.95 x 342.0 g = 666.9 g
  2. Using ratios, find how many g Sucrose associated with 1000 g water
    333 g water associated with 666.9 g Sucrose
    1000 g water associated with 1000 / 333 x 666.9 g = 2002.70 g = 2002.7 to 1 dec pl.
OR:
  1. Using ratios, find how many moles of Sucrose associated with 1000 g water
    333 g water associated with 1.95 moles of Sucrose
    1000 g water associated with 1000 / 333 x 1.95 moles = 5.8558 moles of Sucrose
  2. Find how many g Sucrose associated with 1000 g water
    Mass of Sucrose = 5.8558 x 342.0 g = 2002.68 g = 2002.7 g to 1 dec pl.
EITHER:
  1. Find how many grams of Sucrose are associated with 333 grams of water
  2. Then, using ratios, find how many grams of Sucrose are associated with 1000 grams of water.
OR
  1. Using ratios, find how many moles of Sucrose are associated with 1000 g of water
  2. Find how many grams of Sucrose are associated with 1000 grams of water.

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Question 9 of 9

If 40.5 g of Glucose (RMM 180.0) are associated with 150 g of water, how many moles of Glucose are associated with 1000 g of water?
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moles
1.5 moles of Glucose are associated with 1000 g of water.
EITHER:
  1. Find how many moles of Glucose associated with 150 g water
    Moles of Glucose = 40.5 / 180.0 moles = 0.225 moles
  2. Using ratios, find how many moles associated with 1000 g water
    150 g water associated with 0.225 moles Glucose
    1000 g water associated with 1000 / 150 x 0.225 moles = 1.5 moles Glucose.
OR:
  1. Using ratios, find how many g associated with 1000 g water
    150 g water associated with 40.5 g Glucose
    1000 g water associated with 1000 / 150 x 40.5 g = 270.0 g Glucose
  2. Find how many moles Glucose associated with 1000 g water
    Moles of Glucose = 270.0 / 180.0 moles = 1.5 moles
EITHER:
  1. Find how many moles of Glucose are associated with 150 g water
  2. Using ratios, find how many moles Sucrose associated with 1000 g water
OR
  1. Using ratios, find how many g Sucrose associated with 1000 g water
  2. Find how many moles of Sucrose are associated with 1000 g water.

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