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P(4)S(3)+8O(2) rarr P(4)O(10)+3SO(2) C...

`P_(4)S_(3)+8O_(2) rarr P_(4)O_(10)+3SO_(2)`
Calculate minimum mass of `P_(4)S_(3) is required to produce atleast 1 gm of each product.

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When phosphours (P_(4)) is heated in limited amount of O_(2).P_(4)O_(6) (tetraphosphorous hexaoxide) is obtained, and in excess of O_(2) , P_(4) O_(10) (tetraphosphours decaoxide) is obtained. i. P + 3 O_(2) rarr P_(4) O_(6) , ii. P_(4) + 5O_(2) rarr P_(4)O_(10) What mass of P_(4) O_(6) will be produced by the combustion of 2.0 g of P_(4) with 2.0 g of O_(2) .

P_(4)O_(6) and P_(4)O_(10) are formed by burning P_(4) with O_(2) as: P_(4) + 3O_(2) to P_(4)O_(6) P_(4) + 5O_(2) to P_(4)O_(10) What are the masses of P_(4)O_(6) and P_(4)O_(10) that will be produced by the combustion of 2.0 g of P_(4) in 2.0 g of oxygen leaving no P_(4) and O_(2) ?

The unbalanced equation for the reaction of P_(4)S_(3) with nitrate in aqueous acidic medium is given below. P_(4)S_(3)+NO_(3)^(-)rarr H_(3)PO_(4)+SO_(4)^(2-)+NO The number of moles of water required per mol of P_(4)S_(3) is (x)/(3) , the value of x is

Similar to % labelling of oleum ,a mixture of H_(3)PO_(3) and P_(4)O_(6) is labelled as (100 +x)% where x is maximum mass of water which can reacts with P_(4)O_(6) present in 100 gm mixture of H_(3)PO_(3) and P_(4)O_(6) P_(4)O_(6) + H_(2)O rarr H_(3)PO_(3) For such a mixture of P_(4)O_(6) and H_(3)PO_(3) labelled as (100 +x)% . Value of x can lie in range of (maximum and minimum) :

P_(4)O_(6) reacts with water according to equation P_(4)O_(6)to4H_(3)PO_(3) . Calculate the volume of 0.1MNaOH solution required to neutralise the acid formed by dissolving 1.1g of P_(4)O_(6) in H_(2)O .

Similar to % labelling of oleum ,a mixture of H_(3)PO_(3) and P_(4)O_(6) is labelled as (100 +x)% where x is maximum mass of water which can reacts with P_(4)O_(6) present in 100 gm mixture of H_(3)PO_(3) and P_(4)O_(6) P_(4)O_(6) + H_(2)O rarr H_(3)PO_(3) If such a mixture is labelled as 127 % , then mass of free P_(4)O_(6) in given 100 g mixture is :

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