An aqueous solution of cadmium bromide and an aqueous solution of silver nitrate yields an aqueous solution of cadmium nitrate and solid silver bromide. If you have been given 6.7 L of cadmium bromide and 67 moles of silver nitrate, which reactant is the limiter? silver nitrate is the limiter cadmium bromide is the limiter cadmium nitrate is the limiter silver bromide is the limiter

To determine the limiting reactant, we need to compare the amount of product that can be formed from each reactant.<br /><br />The balanced chemical equation for the reaction is:<br /><br />CdBr2 + 2AgNO3 → Cd(NO3)2 + 2AgBr<br /><br />From the balanced equation, we can see that 1 mole of CdBr2 reacts with 2 moles of AgNO3 to produce 1 mole of Cd(NO3)2 and 2 moles of AgBr.<br /><br />Given that we have 6.7 L of CdBr2, we can calculate the number of moles of CdBr2 using the ideal gas law:<br /><br />PV = nRT<br /><br />Where:<br />P = pressure (assumed to be 1 atm)<br />V = volume (6.7 L)<br />n = number of moles<br />R = ideal gas constant (0.0821 L·atm/mol·K)<br />T = temperature (assumed to be 298 K)<br /><br />n = (PV) / (RT)<br />n = (1 atm × 6.7 L) / (0.0821 L·atm/mol·K × 298 K)<br />n = 0.276 moles of CdBr2<br /><br />Now, we can compare the amount of product that can be formed from each reactant:<br /><br />For CdBr2:<br />0.276 moles of CdBr2 × (1 mole of Cd(NO3)2 / 1 mole of CdBr2) = 0.276 moles of Cd(NO3)2<br /><br />For AgNO3:<br />67 moles of AgNO3 × (1 mole of Cd(NO3)2 / 2 moles of AgNO3) = 33.5 moles of Cd(NO3)2<br /><br />Since the amount of Cd(NO3)2 that can be formed from CdBr2 (0.276 moles) is less than the amount that can be formed from AgNO3 (33.5 moles), CdBr2 is the limiting reactant.<br /><br />Therefore, the correct answer is: cadmium bromide is the limiter.