The number of moles of `Fe_(2)O_(3)` formed when `5.6L` of `O_(2)` reacts with `5.6g` of `Fe` at STP
`0.125`
`0.01`
`0.05`
`0.10`

The oxidation number of Fe in Fe_(0.94)O is

0.5 mole of P_(4)O_(10) contains

The number of moles of H_(2)O_(2) required to completely react with 400 ml of 0.5 N KMnO_(4) in acidic medium are

For the reaction Ba(OH)_(2)+2HClO_(3)rarr Ba(ClO_(3))_(2)+2H_(2)O . Calculate the number of moles of H_(2)O formed when 0.1 mole of Na(OH)_(2) is treated with 0.0250 mole HClO_(3) .

The % of Fe^(+2) in Fe_(0.93)O_(1.00) is .

For the reaction N_(2)+O_(2)hArr 2NO at equilibrium number of moles of N_(2),O_(2) and NO pressent in the system per litre are 0.25 mole, 0.05 moles and 1.0 mole respectively What will be the initial concentration of N_(2) and O_(2) respectively

The mass percentage of Fe^(3+) ion present in Fe_(0.93), O_(1.00) is

What is the number of moles of Fe(OH)_(3) (s) that can be produced by allowing 1 mole of Fe_(2)S_(3) , 2 moles of H_(2)O and 3 moles of O_(2) to react as : 2Fe_(2)S_(3) + 6H_(2)O + 3O_2 rarr 4Fe(OH)_(3) + 6S ?

Calculate the mass of each of the following at STP: 5.6L O_(2)

If 0.5 mol of CaBr_(2) is mixed with 0.2 mol of K_(3) PO_(4) , the maximum nubmer of moles of Ca_(3) (PO_(4))_(2) that can be formed is: a. 0.1 b. 0.2 c. 0.5 d.0.7