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Answer :
Question: What volume of H2 at 765 torr and 25°C is required to reduce 35.5 g of copper (II) oxide?
To determine the volume of H2 gas required to reduce 35.5 g of copper (II) oxide, we can use the concept of stoichiometry and the ideal gas law.
First, let's write the balanced chemical equation for the reaction between H2 gas and copper (II) oxide:
CuO + H2 -> Cu + H2O
From the balanced equation, we can see that one mole of CuO reacts with one mole of H2 to produce one mole of Cu and one mole of H2O.
Next, we need to convert the given mass of CuO to moles. To do this, we use the molar mass of CuO, which is 79.55 g/mol. Thus, the number of moles of CuO can be calculated as:
moles of CuO = mass of CuO / molar mass of CuO
moles of CuO = 35.5 g / 79.55 g/mol
moles of CuO = 0.446 mol
Since the stoichiometric ratio between CuO and H2 is 1:1, the number of moles of H2 required is also 0.446 mol.
Now, we can use the ideal gas law to calculate the volume of H2 gas. The ideal gas law equation is:
PV = nRT
Where:
P is the pressure (765 torr),
V is the volume of the gas,
n is the number of moles of the gas (0.446 mol),
R is the ideal gas constant (0.0821 L·atm/mol·K), and
T is the temperature in Kelvin (25°C + 273.15 = 298.15 K).
Rearranging the equation to solve for V, we have:
V = (nRT) / P
V = (0.446 mol)(0.0821 L·atm/mol·K)(298.15 K) / 765 torr
Converting torr to atm (1 atm = 760 torr), we get:
V = (0.446 mol)(0.0821 L·atm/mol·K)(298.15 K) / (765 torr / 760 torr/atm)
V = 0.017 L or 17 mL
Therefore, the volume of H2 gas required to reduce 35.5 g of copper (II) oxide is approximately 0.017 L or 17 mL.
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