Final answer:
To determine the volume of hydrogen gas produced when 3.00 g of tin reacts with excess HF, we need to use stoichiometry. First, convert the mass of tin to moles. Then, use the mole ratio to determine the moles of hydrogen gas produced. Finally, use the density of hydrogen gas to convert moles to volume.
Step-by-step explanation:
To determine the volume of hydrogen gas produced when 3.00 g of tin reacts with excess HF, we need to use stoichiometry. First, we convert the mass of tin to moles by dividing it by the molar mass of tin (118.71 g/mol). This gives us 0.0253 moles of tin. Next, we use the balanced equation to determine the mole ratio between tin and hydrogen gas, which is 1:1. Therefore, we have 0.0253 moles of hydrogen gas. Finally, we use the density of hydrogen gas (0.09 g/L) to convert moles to volume. Since the density is given in grams per liter, we can multiply the moles of hydrogen gas by the molar mass of hydrogen gas (2.02 g/mol) to find the mass of hydrogen gas produced. Then, we divide the mass by the density to get the volume in liters. Multiplying by 1000, we convert it to milliliters.
0.0253 mol H2 x 2.02 g H2/mol = 0.0511 g H2
0.0511 g H2 / 0.09 g/L = 0.567 L H2
0.567 L H2 x 1000 mL/L = 567 mL H2