Question

Question #b786d

Answer 1

0,01 mol of CO and 0,03 mol of `CO_2`

Explanation:

You can get the n° di mol of the mixture of CO and `CO_2` from the gases' law where one mol occupy at NTP ( STP are 1 atm and 273 K, but NTP is less sure) 22,4 L. `n= PV/RT = (1 atm xx 0,896 L)/ (0,082(L xx Atm)/(mol xx K) xx 273 K)=0,0400 mol`

Now you must resolve a system;
if you call x the mol of CO and Y the mol of `CO_2` and knowing that the weigh of each gas is given by the mol of gas multiplyed for the MM you have:
X+Y = 0,04 and
`X xx 28 + Y xx 44 = 1,28 g`.
From the first equation you have X= 0,04 -Y and in the second
1,12 + (44-28) Y = 1,28 ;
Y = (1,28-1,12)/16 = 0,01 mol of CO and then 0,03 mol of `CO_2`