How many grams of AgCl (molar mass = 143.5g/mol) can be produced when 25.0ml of 1.5M AgNO₃ completely reacts with excess NaCl(s)? AgNO₃(aq) + NaCl(s) --> AgCl(s) + NaNO₃(aq)

To calculate the grams of AgCl produced, we need to use stoichiometry and the molarity of AgNO₃. From the balanced equation, 1 mole of AgNO₃ reacts with 1 mole of AgCl. The moles of AgCl produced is equal to the moles of AgNO₃, which can be calculated by multiplying the volume (in L) of AgNO₃ by its molarity. This gives us 0.0375 moles of AgCl, which, when multiplied by its molar mass (143.5 g/mol), results in 5.36 grams of AgCl.

Explanation:

To calculate the grams of AgCl produced, we need to use stoichiometry and the molarity of AgNO₃. From the balanced equation, we can see that 1 mole of AgNO₃ reacts with 1 mole of AgCl. So the moles of AgCl produced is equal to the moles of AgNO₃. The moles of AgNO₃ can be calculated by multiplying the volume (in L) of AgNO₃ by its molarity.

The volume of AgNO₃ is given as 25.0 ml (which is the same as 0.025 L) and the molarity is given as 1.5 M. Multiplying these two values together gives us 0.0375 moles of AgNO₃.

Since the moles of AgCl produced is equal to the moles of AgNO₃, we have 0.0375 moles of AgCl. To calculate the grams of AgCl, we use the molar mass of AgCl. Multiplying the moles of AgCl by its molar mass (143.5 g/mol) gives us the answer: 5.36 grams of AgCl.

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