A 0.094 g sample of zinc metal is dissolved in excess 0.1 M HCl. What volume of H2 is produced at 273.15 K, 1 atm? R = 0.082057 L atm K-1mol-1.Group of answer choices0.064 L0.032 L0.045 L0.025 L0.016 L
Question
A 0.094 g sample of zinc metal is dissolved in excess 0.1 M HCl. What volume of H2 is produced at 273.15 K, 1 atm? R = 0.082057 L atm K-1mol-1.Group of answer choices0.064 L0.032 L0.045 L0.025 L0.016 L
Solution
To solve this problem, we need to follow these steps:
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First, we need to convert the mass of zinc to moles. The molar mass of zinc is approximately 65.38 g/mol. So, 0.094 g of zinc is 0.094/65.38 = 0.00144 moles.
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The balanced chemical equation for the reaction of zinc with hydrochloric acid is: Zn(s) + 2HCl(aq) -> ZnCl2(aq) + H2(g). From this equation, we can see that one mole of zinc produces one mole of hydrogen gas.
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Therefore, 0.00144 moles of zinc will produce 0.00144 moles of hydrogen gas.
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Now, we can use the ideal gas law to find the volume of hydrogen gas. The ideal gas law is PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature. We are solving for V, so we rearrange the equation to V = nRT/P.
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Substituting the given values into the equation gives us V = (0.00144 moles)(0.082057 L atm K-1mol-1)(273.15 K) / 1 atm = 0.032 L.
So, the volume of H2 produced is 0.032 L.
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