Q.What weights of `P_4O_6` and `P_4O_10` will be produced by the combustion of 2 g of `P_4` in 2 g of oxygen leaving no `P_4` and `O_2` ?

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The combustion of phosphorus `P_4` with oxygen `O_2` produces two oxides `P_4O_6` and `P_4O_10` .

The molar mass of `P_4` is 124. 2 gram of `P_4` is equivalent to `2/124 ` moles of `P_4` . The molar mass of `O_2` is 32, 2 g of `O_2` is equivalent to `2/32` moles of `O_2` .

Let the number of moles of `P_4O_6` formed be X and the number of moles of `P_4O_10` formed be equal to Y.

X + Y = `2/124` and 3*X + 5*Y = `2/32`

Substituting `X = 2/124 - Y` in `3*X + 5*Y = 2/32` gives `3*(2/124 - Y) + 5*Y = 2/32`

=> `2Y = 2/32 - 3/62 `

=> `Y = 7/992`

X = `9/992`

The mass of `9/992` moles of `P_4O_6` is `495/248` g and the mass of `7/992` moles of `P_4O_10` is `497/248` g

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