A steel container with a volume of 500.0 mL is evacuated, and 25.0 g of CaCO₃ is added. The container and contents are then heated to 1500 K, causing the CaCO₃ to decompose completely, according to the equation CaCO₃(s) → CaO(s) + CO₂(g). (a) Using the ideal gas law and ignoring the volume of any solids remaining in the container, calculate the pressure inside the container at 1500 K.

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Hi everyone here we have a problem telling us that the following figure shows 23.50 liter vessels at 27.0 degrees Celsius that are connected by a stop cock. The vessel on the left contains 7.50 g of ammonia gasses, and the vessel on the right contains 7.50 g of hydrogen bromide grasses. Once the stop cock is open, the two gasses will react as shown below ammonia gasses plus hydrogen bromide, gaseous forms ammonium bromide determine the final pressure of the setup. Once the reaction is complete, the volume occupied by ammonium bromide solid can be ignored. So the molar mass of ammonia is 17 . grams per mole. The moles of ammonia bromide Equals 7.50 g of ammonia times one mole of ammonia, divided by 17.034g of ammonia Equals 0.4403 moles. The molar mass of hydrogen bromide equals .912 g per mole. The moles of ammonium bromide Equals 7.50 g of hydrogen bromide times one mole of hydrogen bromide divided by 80.912g of hydrogen bromide Equals 0.09269 M.nolds. So now we need to determine the limiting reactant. So we need to calculate the moles of product. Each reactant can produce. So the moles of ammonium bromide Equals 0. moles of ammonia times one mole of ammonium bromide over one mole of ammonia Equals 0. moles of ammonium bromide. The moles of ammonium bromide equals zero .09269 moles times one mole of ammonium bromide, over one mole of hydrogen bromide Equals 0.0927 moles of ammonium bromide. So this is our limiting reactant, which means it is completely consumed. So now we're going to calculate our molds of ammonia lift. Are moles of ammonia reacted Equals 0.0927 ammonium bromide times one mole ammonia over one mole ammonium bromide Equals 0.092, seven moles of ammonia reacted. So our ammonia lift Equals 0. moles -0.0927 moles Equals 0. moles. And those are our answers. Thank you for watching. Bye.

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Key Concepts

Ideal Gas Law

Stoichiometry of Decomposition Reactions

Gas Behavior at High Temperatures