1) Into a 0.25 M solution of NaCl(aq), excess Al(NO3)3(aq) was added to form AlC

Question

1) Into a 0.25 M solution of NaCl(aq), excess Al(NO3)3(aq) was added to form AlCl3(s). 3NaCl(aq) + Al(NO3)3(aq) ? AlCl3(s) + 3NaNO3(aq) If you knew the volume of the solution containing NaCl(aq), determine how you would predict the mass of AlCl3(s) formed by completing the following solution map. Drag the appropriate labels to their respective targets.

2)Dimensional analysis simply refers to the inclusion of units in an equation. After setting up the solution map, dimensional analysis can be used to set up the conversion factors that lead to the desired values within this map. The correct setup of an equation can be verified by checking if the result will only have the desired units after unit cancellation. 3NaCl(aq) + Al(NO3)3(aq) ? AlCl3(s) + 3NaNO3(aq) For the described reaction, you have a 0.280 L solution of 0.700 M NaCl(aq). Using the solution map from Part A as a guide, complete the following dimensional analysis that allows you to calculate the moles of AlCl3(s) that can be formed by placing the values of each conversion factor according to whether they should appear in the numerator or denominator. Drag the appropriate values to their respective targets.

3) When you are presented with additional details, it can sometimes be a little more difficult to determine which information is necessary and how molarity should be applied. Using both a solution map and dimensional analysis can help clarify which information and conversion factors are necessary to determine the desired value. The following dimensional analysis setup could be used to determine the theoretical mass of AlBr3(s) (molecular mass = 266.69 g/mol ) produced based on reacting 62.2 g of a 0.016 mol/L solution of Br2(l) (density = 1016 g/L ) with excess Al(s) as described in the following equation: 3Br2(l) + 2Al(s) ? 2AlBr3(s) Complete the dimensional analysis for calculating the mass of the product by placing the values of each conversion factor according to whether they should appear in the numerator or denominator when calculating the mass of AlBr3(s) produced from a sample of Br2(l). Drag the appropriate values to their respective targets.

Explanation / Answer

1) The reaction here is: 3NaCl(aq) + Al(NO3)3(aq) -----> AlCl3(s) + 3NaNO3(aq)

For any substance given we can use the two formula below to find out the missing information.

For Solids: Moles = Mass/Molar mass ---------1

For Liquids: Moles = Molarity * Volume ---------2

Amongst the two reactants, volume and molarity of NaCl are given. So we can find moles of NaCl using formula 2. Moles NaCl = Molarity NaCl * Volume of NaCl.

Once we have moles of NaCl, using the molar ratio we can find out moles of AlCl3(s). (we are not considering Al(NO3)3 for comaprison as it is taken in excess). For every 3 moles of NaCl, 1 mole of AlCl3 is produced.

Once we have moles of AlCl3 we can calulate mass of AlCl3 using the formula 1.

Mass of AlCl3 = Moles AlCl3 * Molar mass of AlCl3

The solution map would be

Vol NaCl (x molarity of NaCl) moles NaCl (divide by 3) moles AlCl3 (x molarmass of AlCl3) mass AlCl3

2) Using the solution map above dimensional analysis can be set as follows:

0.280L NaCl x (0.700mol NaCl/1L NaCl) x (1mol AlCl3/ 3mol NaCl) x (133.34 gms AlCl3/1mol AlCl3) = 8.71 gms AlCl3

3) The reaction involved here is: 3Br2 (l) + 2Al (s) -----> 2AlBr3 (s)

Solution Map: mass  Br2 -----> Vol Br2 ---> moles Br2 ---> moles of AlBr3 ---> mass of AlBr3

Solution: moles of Br2 = mass of Br2 x (1/density of Br2 ) x Molarity of Br2

Dimensional Analysis:

62.2g Br2x(1L Br2/1016g Br2)x(0.016mol Br2/1.0L Br2)x(2mol AlBr3/3mol Br2)x(266.69g AlBr3/1mol AlBr3) = 0.174 g AlBr3

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