Question

According to the following equation what volume of HCl(g) measured at STP can be produced from 6.65 g of H2 and excess Cl2?

H2(g) + Cl2(g) → 2HCl(g)?


22.4 L

74.5 L

298 L

149 L

Answer 1

Answer: 149 L

Explanation: since it's measured at stp, you don't have to worry about changing the temperature or the pressure. in this case, you're going to take the 6.65 g of H2 and figure out the molecular weight for it. it'll be around 2, which is the number I'm going to use. this allows us to transfer H2 from grams to moles. then you'll look at the balanced equation. for every 1 mole of H2, there's 2 moles of HCL, which is where we're trying to get. we'll add that ratio to move from H2 to HCL. then, we'll just transition it from moles of HCL to liters by using the stp conversion of 22.4. solving the equation, you get 149.