6. How many grams of barium chloride are produced when 1.25times 10^23 molecules of HCl reacts with barium according to the following equation? Word equation: hydrochloric acid+bariumarrow barium hydroxide+hydrogen Formula equation: 2HCl+1Baarrow 1BaCl_(2)+1H_(2)

To solve this problem, we need to follow these steps:<br /><br />1. Convert the given number of molecules of HCl to moles.<br />2. Use the stoichiometry of the balanced chemical equation to determine the moles of BaCl2 produced.<br />3. Convert the moles of BaCl2 to grams.<br /><br />Step 1: Convert the given number of molecules of HCl to moles.<br />Given: $1.25\times 10^{23}$ molecules of HCl<br />Moles of HCl = $\frac{1.25\times 10^{23}}{6.022\times 10^{23}}$ = 2.07 moles<br /><br />Step 2: Use the stoichiometry of the balanced chemical equation to determine the moles of BaCl2 produced.<br />According to the balanced chemical equation, 2 moles of HCl produce 1 mole of BaCl2.<br />Moles of BaCl2 produced = $\frac{2.07}{2}$ = 1.035 moles<br /><br />Step 3: Convert the moles of BaCl2 to grams.<br />Molar mass of BaCl2 = 208.23 g/mol<br />Mass of BaCl2 produced = 1.035 moles × 208.23 g/mol = 214.6 grams<br /><br />Therefore, 214.6 grams of barium chloride are produced when $1.25\times 10^{23}$ molecules of HCl react with barium according to the given equation.