To calculate the pressure of the oxygen gas (\( O_2 \)) produced from heating silver(I) oxide (\( AgO \)), we can follow these steps:
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Determine the reaction and products: When \( AgO \) is heated, it decomposes into silver (\( Ag \)) and oxygen gas (\( O_2 \)): \[ 2AgO(s) \rightarrow 2Ag(s) + O_2(g) \]
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Calculate moles of \( AgO \): First, find the molar mass of \( AgO \):
- \( Ag \): approximately \( 107.87 , g/mol \)
- \( O \): approximately \( 16.00 , g/mol \)
- Molar mass of \( AgO = 107.87 + 16.00 = 123.87 , g/mol \)
Now calculate the number of moles of \( AgO \): \[ \text{moles of } AgO = \frac{5.76 , g}{123.87 , g/mol} \approx 0.0464 , mol \]
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Determine moles of \( O_2 \) produced: According to the balanced equation, from 2 moles of \( AgO \), 1 mole of \( O_2 \) is produced. Thus: \[ \text{moles of } O_2 = \frac{0.0464 , mol , AgO}{2} \approx 0.0232 , mol \]
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Use the Ideal Gas Law to calculate pressure: The Ideal Gas Law is given by: \[ PV = nRT \] Where:
- \( P \) = pressure (in atm)
- \( V \) = volume (in L)
- \( n \) = number of moles of gas
- \( R \) = ideal gas constant \( 0.0821 , L \cdot atm/(K \cdot mol) \)
- \( T \) = temperature (in K, needs to be provided or assumed)
Assuming the temperature is given, let’s use \( T = 298 , K \) (which is room temperature): \[ P = \frac{nRT}{V} \] Plugging in the values: \[ P = \frac{(0.0232 , mol)(0.0821 , L \cdot atm/(K \cdot mol))(298 , K)}{0.65 , L} \] \[ P \approx \frac{0.5704 , L \cdot atm}{0.65 , L} \approx 0.876 , atm \]
So, the pressure of the \( O_2 \) gas in the flask under these assumptions (with assumed temperature) is approximately 0.88 atm. If the temperature is different, please provide that value for a more accurate calculation.
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