A mixture of 1.0 mole of Al and 3.0 mole of `Cl_(2)` are allowed to react as:
`2Al(s)+3Cl_(2)(g) to 2AlCl_(3)(g)`. Then moles of excess reagent left unreacted is:

A mixture of 1 mol of Al and 3 mol of Cl_2 are allowed to react as 2Al_(s) + 3Cl_(2(g)) rarr 2AlCl_3 How many moles of excess reactant left unreacted?

A mixture of 1 mol of Al and 3 mol of Cl_2 are allowed to react as 2Al_(s) + 3Cl_(2(g)) rarr 2AlCl_3 How many moles of AlCl_3 are formed?

A mixture of 1 mol of Al and 3 mol of Cl_2 are allowed to react as 2Al_(s) + 3Cl_(2(g)) rarr 2AlCl_3 Identify the limiting reagent.

A mixture of 1.0 mole of Al and 3.0 mole of Cl_(2) are allowed to react as: 2Al_((s))+3Cl_(2) rarr 2AlCl_(3(S)) (a) Which is limiting reagent? (b) How many moles of AlCl_(2) are formed? (c ) Moles of excess reagent left unreacted.

A mixture of 0.3 mole of H_(2) and 0.3 mole of I_(2) is allowed to react in a 10 lit vessel at 500^(@)C . If K_(C) of H_(2) +I_(2)

A mixture of 0.3 moles of H_2 and 0.3 moles of l_2 is allowed to react in a 10 litre evacuated flask at 500°C. The reaction is H_2 + I_2 = 2HI, the K is found to be 64. The amount of unreacted 'l_2' equilibrium is

Initially 0.8 mole of PCl_(5) and 0.2 mole of PCl_(3) are mixed in one litre vessel . At equilibrium 0.4 mole of PCl_(3) is present . The value of K_(C) for the reaction PCl_(5(g)) hArr PCl_(3(g)) + Cl_(2(g)) is