# Question #b786d

Jul 28, 2017

0,01 mol of CO and 0,03 mol of $C {O}_{2}$

#### Explanation:

You can get the n° di mol of the mixture of CO and $C {O}_{2}$ 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. $n = P \frac{V}{R} T = \frac{1 a t m \times 0 , 896 L}{0 , 082 \frac{L \times A t m}{m o l \times K} \times 273 K} = 0 , 0400 m o l$

Now you must resolve a system;
if you call x the mol of CO and Y the mol of $C {O}_{2}$ 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
$X \times 28 + Y \times 44 = 1 , 28 g$.
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 $C {O}_{2}$