In the chemical reaction 2CO(g) + O2 —-> 2CO2 if you react 55.0g of oxygen gas you would expect to produce what volume of carbon dioxide at stp

13mL of CO₂ are produced

Explanation:

Based on the chemical reaction:

2CO(g) + O₂(g) → 2CO₂(g)

Where 1 mole of O₂ in excess of CO produce 2 moles of CO₂}

Thus, to solve this problem we need to convert the mass of oxygen gas to moles and, using the chemical reaction we can find the moles of CO₂ and its volume at STP:

Moles O₂ -Molar mass: 32g/mol-

55g O₂ * (1mol / 32g) = 1.72 moles O₂

Moles CO₂:

1.72 moles O₂ * (2 moles CO₂ / 1mol O₂) = 3.44 moles CO₂

Volume:

PV = nRT

V = P / nRT

Where P is pressure = 1atm at STP

n are moles = 3.44 moles of CO₂

R is gas constant = 0.082atmL/molK

T is absolute temperature = 273.15K at STP

V = 1atm / 3.44mol*0.082atmL/molK*273.15K

V = 0.0130L = 13mL of CO₂ are produced

just did the assignment and the answer was #1 +2 n

the answer is barium and copper bromide