Question

If 56.0g of N2 gas occupies 44.8L under standard conditions, then what is the mass of 134.4L of H2 gas under the same conditions?

What mass of NH3 will be formed by the reaction between the two gases above? Write the complete balanced reaction first.

Answer 1

12.09528g
`H_(2)`

68.12208g
`NH_(3)`

Step-by-step explanation:

`n = (V)/(V_(m) ) \\ = (134.4)/(22.4) \\ = 6 mol H_(2)`

`n=(m)/(M) \\m= nM\\=(6)(2.051588)\\=12.09528g H_(2)`

`N_(2) : NH_(3) \\ 1 : 2\\2:x\\x= 4mol NH_(3) \\\\n = (m)/(M) \\m=nM\\=(4)(17.03052)\\=68.12208g NH_(3)`