Answer:
5.38 g of CO.
Step-by-step explanation:
From the question given above, the following data were obtained:
Volume of CO = 4.3 L
Mass of CO =?
Next, we shall determine the number of mole of CO that occupied 4.3 L at STP. This can be obtained as follow:
Recall = 1 mole of any gas occupy 22.4 L at STP.
1 mole of CO occupied 22.4 L at STP.
Therefore, Xmol of CO will occupy 4.3 L at STP i.e
Xmol of CO = 4.3 / 22.4
Xmol of CO = 0.192 mole.
Thus, 0.192 mole of CO occupy 4.3 L at STP.
Finally, we shall determine the mass of CO as follow:
Mole of CO = 0.192 mole.
Molar mass of CO = 12 + 16 = 28 g/mol
Mass of CO =?
Mole = mass /Molar mass
0.192 = mass of CO / 28
Cross multiply
Mass of CO = 0.192 × 28
Mass of CO = 5.38 g.
Therefore, 5.38 g of CO occupied 4.3 L at STP.