5) 3.2g of barium chloride reac] with 2.2g of lithium bromide Determine the moles of lithium chloride and identify the limiting and excess reactants. __ __

## Step 1<br />First, we need to calculate the molar mass of each reactant. The molar mass of a compound is the sum of the molar masses of its constituent atoms.<br /><br />### The molar mass of barium chloride (BaCl2) is calculated as follows:<br />\(Molar\, mass\, of\, BaCl2 = 137.33 (Ba) + 2*35.45 (Cl) = 208.23 g/mol\)<br /><br />### The molar mass of lithium bromide (LiBr) is calculated as follows:<br />\(Molar\, mass\, of\, LiBr = 6.94 (Li) + 79.90 (Br) = 86.84 g/mol\)<br /><br />## Step 2<br />Next, we calculate the number of moles of each reactant. The number of moles is calculated by dividing the mass of the substance by its molar mass.<br /><br />### The number of moles of BaCl2 is calculated as follows:<br />\(Moles\, of\, BaCl2 = \frac{3.2g}{208.23g/mol} = 0.0154 mol\)<br /><br />### The number of moles of LiBr is calculated as follows:<br />\(Moles\, of\, LiBr = \frac{2.2g}{86.84g/mol} = 0.0253 mol\)<br /><br />## Step 3<br />We then compare the number of moles of each reactant to determine which one is the limiting reactant. The limiting reactant is the one that is completely consumed in the reaction and determines the maximum amount of product that can be formed.<br /><br />### In this case, BaCl2 is the limiting reactant because it has fewer moles (0.0154 mol) than LiBr (0.0253 mol).<br /><br />## Step 4<br />Finally, we calculate the number of moles of lithium chloride (LiCl) that can be formed from the limiting reactant. The stoichiometry of the reaction tells us that one mole of BaCl2 reacts with one mole of LiBr to produce one mole of LiCl.<br /><br />### Therefore, the number of moles of LiCl that can be formed is equal to the number of moles of the limiting reactant, BaCl2, which is 0.0154 mol.