What volume of H₂, in milliliters, at 21 °C and 0.998 atm can be produced from the reaction of 2.40 grams of Zn in excess HCl?
Zn(s) + 2 HCl(aq) $\rightarrow$ ZnCl₂(aq) + H₂(g)

Question

What volume of H₂, in milliliters, at 21 °C and 0.998 atm can be produced from the reaction of 2.40 grams of Zn in excess HCl? Zn(s) + 2 HCl(aq) $\rightarrow$ ZnCl₂(aq) + H₂(g)

Quizplus · Accepted Answer

Using the ideal gas law and stoichiometry, 2.40 g of Zn produces approximately 0.0367 moles of H2. At 21 °C and 0.998 atm, this corresponds to about 890 mL of H2 gas.

What Volume of H2, in Milliliters, at 21 °C and 0.998

What Volume of H₂, in Milliliters, at 21 °C and 0.998