Question

896ml of a mixture of CO and CO2 weigh 1.28g at NTP.Calculate the volume of CO2 in the mixture at NTP

Answer 1

Hey!
Let's start by assuming a mixture in which
V(CO2) = V mL, then automatically V(CO) = (896 - V) mL

At NTP (298 K and 1 atm)
Number of moles of CO2=
$n(co2) = \frac{v}{22400} = \frac{w(co2)}{44}$
where w(CO2) is the weight of CO2 in grams.
From here we get,
w(CO2) = 44V/22400 - (1)

Similarly for CO, we get,
w(CO) = 28(896 - V)/ 22400 - (2)

Now, we know that w(CO) + w(CO2) = 1.28 g
{44V + 25088 - 28V} / 22400 = 1.28
{44V + 25088 - 28V} = 28672
16 V = 3584
V = 224 mL

Answer 2

Answer:

ans will be 224 ml

Explanation:

please look at pics....hope you understood...