Question

An excess of na (s) is added to 100 ml of 0.8 m hcl at stp. What volume of h2 gas can be produced?

Answer 1

The volume of H2 gas produced when an excess of sodium reacts with 100 ml of 0.8 M HCl at STP will be 896 mL.

Step-by-step explanation:

The volume of H2 gas produced when an excess of sodium (Na) reacts with 100 ml of 0.8 M HCl can be calculated using stoichiometry. The balanced chemical equation for the reaction is:

2 Na (s) + 2 HCl (aq) → 2 NaCl (aq) + H2 (g)

First, we must determine the number of moles of HCl in the solution:

0.1 L × 0.8 mol/L = 0.08 mol HCl

Since the reaction ratio between Na and HCl is 2:2, the same amount of HCl moles will produce half the number of moles of H2 because of the 2:1 ratio in the reaction for HCl to H2. Therefore, 0.08 mol of HCl will produce 0.04 mol of H2 gas.

At STP, 1 mole of gas occupies 22.4 liters, so 0.04 mol will occupy:

0.04 mol × 22.4 L/mol = 0.896 L of H2

So, the volume of H2 gas produced at STP will be 896 mL.

Answer 2

The volume of hydrogen gas that can be produced when sodium (Na) reacts with 0.8 M hydrochloric acid (HCl) at STP is 1.792 liters.

Step-by-step explanation:

The student is asking how to calculate the volume of hydrogen gas that can be produced when sodium (Na) reacts with hydrochloric acid (HCl) at standard temperature and pressure (STP), which is 0 degrees Celsius and 1 atm of pressure. The reaction between sodium and hydrochloric acid is as follows:

2 Na(s) + 2 HCl(aq) → 2 NaCl(aq) + H₂(g)

First, we need to calculate the number of moles of HCl using the volume and molarity provided:

0.100 L * 0.8 mol/L = 0.08 mol HCl

Since the reaction ratio of Na to HCl is 1:1, there will also be 0.08 moles of H2 produced. At STP, 1 mole of any gas occupies 22.4 liters, so:

0.08 mol * 22.4 L/mol = 1.792 L

Thus, 1.792 liters of H2 gas can be produced at STP.