- Download PDF
1 Answer | Add Yours
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
We’ve answered 327,775 questions. We can answer yours, too.Ask a question