On mixing 56 gm of CaO with 63 gm of HNO₃, the amount of Ca(NO₃)₂ formed is __________ gm.

ANS:A - 82

To solve this problem, we need to determine the limiting reactant between calcium oxide (CaO) and nitric acid (HNO3) and then use stoichiometry to find the amount of calcium nitrate (Ca(NO3)2) formed. First, we need to find the moles of each reactant:

1. Moles of CaO: Moles of CaO=Mass of CaOMolar mass of CaO=56 g40.08 g/mol=1.396 molMoles of CaO=Molar mass of CaOMass of CaO​=40.08g/mol56g​=1.396mol
2. Moles of HNO3: Moles of HNO3=Mass of HNO3Molar mass of HNO3=63 g63.02 g/mol=0.999 molMoles of HNO3=Molar mass of HNO3Mass of HNO3​=63.02g/mol63g​=0.999mol

Next, we need to determine the stoichiometric ratio between CaO and HNO3: The balanced chemical equation for the reaction between CaO and HNO3 is: CaO+2HNO3→Ca(NO3)2+H2OCaO+2HNO3​→Ca(NO3​)2​+H2​O From the balanced equation, we see that 1 mole of CaO reacts with 2 moles of HNO3 to produce 1 mole of Ca(NO3)2. Now, let's determine the limiting reactant. The limiting reactant is the one that is completely consumed in the reaction and determines the amount of product formed. Since 1 mole of CaO reacts with 2 moles of HNO3, but we only have 0.999 moles of HNO3, HNO3 is the limiting reactant. Now, let's use stoichiometry to find the amount of Ca(NO3)2 formed: From the balanced equation, we see that 1 mole of Ca(NO3)2 is formed for every 2 moles of HNO3 reacted