6.72 L of Carbon Monoxide (CO) is needed.
Explanation:
First from the mass of the oxygen in grams is converted into moles and then it is proportionate to the moles of CO and then it is converted into its Volume in litres ( L).
4.8 g of O₂ × 1 mol of O₂× 2 mol CO ×22.4 L CO / (32 g of O₂ × 1 mol of O₂ × 1 mol of CO) = 6.72 L CO
The volume of CO reacted with oxygen at STP has been 6.72 L.
The moles of reactant and product in a balanced chemical equation has been given by the stoichiometric coefficient.
The balanced chemical equation for the formation of carbon dioxide has been:
[tex]\rm 2\;CO\;+\;O_2\;\rightarrow\;2\;CO_2[/tex]
The balanced chemical equation has been giving the information of 2 moles of CO reacted with a mole of oxygen.
The available moles of oxygen in 4.80 g has been:
[tex]\rm Moles\;=\;\dfrac{Mass}{Molar\;mass}\\\\ Moles\;O_2=\dfrac{4.80}{32}\\\\ Moles\;O_2=0.15\;mol[/tex]
The available moles of oxygen has been 0.15 mol.
The moles of CO reacted has been given by the stoichiometric coefficient of the balanced equation as:
[tex]\rm 1\;mol\;O_2=2\;mol\;CO\\0.15\;mol\;O_2=0.30\;mol\;CO[/tex]
The moles of CO reacted has been 0.30 mol.
The volume of a mole of gas at STP has been 22.4 L. The volume of 0.30 mol CO has been:
[tex]\rm 1\;mol=22.4\;L\\\\0.30\;mol=\dfrac{22.4}{1}\;\times\;0.30\;L\\\\ 0.30\;mol=6.72\;L[/tex]
The volume of CO reacted with oxygen has been 6.72 L.
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