starting with 1.342 g Zn and excess HCL, the volume of H2 gas collected was 472

Question

starting with 1.342 g Zn and excess HCL, the volume of H2 gas collected was 472 mL. the atmospheric pressure was 754 mmHg and the temperature was 21 degrees C. write a balanced equation and calculate the moles of h2 produced by the reaction.

Explanation / Answer

tarting with 1.342 g Zn and excess HCL, the volume of H2 gas collected was 472 mL. the atmospheric pressure was 754 mmHg and the temperature was 21 degrees C. write a balanced equation and calculate the moles of h2 produced by the reaction. ANSWER: Gas-Stoichiometry Porblem. Step 1: Balance the equation. Zn + 2 HCl -------> H2 + ZnCl2 Step 2: Convert the Zn mass to moles (Atomic mass of Zn = 65.41 g/mol). Moles of Zn = 1.342 g/65.41 g/mol = 0.0205 moles of Zn. Step 3: Use of Zn mass Convert Zn moles to H2 moles (Moles ratio of Zn:H2 is 1:1). 0.0205 moles of Zn x(1Mole of H2 /1Mole of Zn) = 0.0205 moles of H2. [OR ] Step 3: Use of the data of H2. Ideal Gas formula : PV = nRT. n (moles) = PV/RT Where P = Pressure =754 mmHg or (754/760)=0.992 atm V = Volume = 472 ml or 0.472 L. T = Temperature = 21 C or 294 K R = Gas constant = 0.0821 atm L/mol K Volume = ((0.992 atm)x(0.472L)/(0.0821 atm L/mol K)x(294 K)] =0.0194 moles of H2 So the moles of H2 gas is 0.0194 moles.

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