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10 mL of hydrogen combines with 5 mL of ...

`10 mL` of hydrogen combines with `5 mL` of oxygen to yield water. When `200 mL` of hydrogen at `STP` is passed over heated `CuO`, the `CuO` loses `0.144 g` of its wieght. Do these result correspond to the law of constant composition?

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To determine whether the results correspond to the law of constant composition, we will analyze both cases step by step. ### Step 1: Calculate the moles of hydrogen (H₂) in the first case Given: - Volume of hydrogen (H₂) = 10 mL - At STP, 1 mole of gas occupies 22,400 mL. To find the moles of H₂: \[ \text{Moles of H₂} = \frac{\text{Volume of H₂}}{\text{Molar volume at STP}} = \frac{10 \, \text{mL}}{22400 \, \text{mL/mol}} = \frac{10}{22400} \, \text{mol} \] ### Step 2: Calculate the mass of hydrogen (H₂) The molar mass of hydrogen (H₂) is 2 g/mol. \[ \text{Mass of H₂} = \text{Moles of H₂} \times \text{Molar mass of H₂} = \left(\frac{10}{22400}\right) \times 2 \, \text{g} = \frac{20}{22400} \, \text{g} = 0.000892857 \, \text{g} \] ### Step 3: Calculate the moles of oxygen (O₂) in the first case Given: - Volume of oxygen (O₂) = 5 mL To find the moles of O₂: \[ \text{Moles of O₂} = \frac{\text{Volume of O₂}}{\text{Molar volume at STP}} = \frac{5 \, \text{mL}}{22400 \, \text{mL/mol}} = \frac{5}{22400} \, \text{mol} \] ### Step 4: Calculate the mass of oxygen (O₂) The molar mass of oxygen (O₂) is 32 g/mol. \[ \text{Mass of O₂} = \text{Moles of O₂} \times \text{Molar mass of O₂} = \left(\frac{5}{22400}\right) \times 32 \, \text{g} = \frac{160}{22400} \, \text{g} = 0.007142857 \, \text{g} \] ### Step 5: Calculate the mass ratio of H₂ to O₂ in the first case \[ \text{Mass ratio} = \frac{\text{Mass of H₂}}{\text{Mass of O₂}} = \frac{0.000892857 \, \text{g}}{0.007142857 \, \text{g}} = \frac{1}{8} \] ### Step 6: Analyze the second case with 200 mL of H₂ Given: - Volume of hydrogen (H₂) = 200 mL To find the moles of H₂: \[ \text{Moles of H₂} = \frac{200 \, \text{mL}}{22400 \, \text{mL/mol}} = \frac{200}{22400} \, \text{mol} \] ### Step 7: Calculate the mass of hydrogen (H₂) in the second case \[ \text{Mass of H₂} = \text{Moles of H₂} \times \text{Molar mass of H₂} = \left(\frac{200}{22400}\right) \times 2 \, \text{g} = \frac{400}{22400} \, \text{g} = 0.017857143 \, \text{g} \] ### Step 8: Use the mass loss of CuO to find the mass of oxygen (O₂) Given: - Mass loss of CuO = 0.144 g (which corresponds to the mass of O₂) ### Step 9: Calculate the mass ratio of H₂ to O₂ in the second case \[ \text{Mass ratio} = \frac{\text{Mass of H₂}}{\text{Mass of O₂}} = \frac{0.017857143 \, \text{g}}{0.144 \, \text{g}} \approx \frac{1}{8} \] ### Conclusion In both cases, the mass ratio of hydrogen to oxygen is consistently 1:8. Therefore, these results correspond to the law of constant composition, which states that a chemical compound always contains its component elements in fixed ratio by mass.
To determine whether the results correspond to the law of constant composition, we will analyze both cases step by step. ### Step 1: Calculate the moles of hydrogen (H₂) in the first case Given: - Volume of hydrogen (H₂) = 10 mL - At STP, 1 mole of gas occupies 22,400 mL. To find the moles of H₂: ...
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