How many liters of hydrogen at STP will be produced when 2.36 grams of magnesium react? Mg(s) + 2HNO3(aq) → Mg(NO3)2(aq) + H2(g)

Answer 1

Mg(s) + 2HNO₃(aq) → Mg(NO₃)₂(aq) + H₂(g) .. balanced equation

moles Mg = 2.36 g x 1 mol Mg / 24.3 g = 0.0971 mols Mg

moles H₂ gas formed = 0.0971 mol Mg x 1 mol H₂ / mol Mg = 0.0971 mols H₂

At STP, 1 mole of any ideal gas will occupy 22.414 liters. Thus...

Liters H₂ gas = 0.0971 mols H2 x 22.414 L / mol = 2.18 liters

Answer 2

To determine the volume of hydrogen gas produced at STP (Standard Temperature and Pressure) when 2.36 grams of magnesium (Mg) reacts, we need to use stoichiometry and the ideal gas law.

First, let's calculate the number of moles of magnesium using its molar mass:

Molar mass of Mg = 24.3 g/mol

Moles of Mg = 2.36 g / 24.3 g/mol ≈ 0.097 moles

From the balanced equation, we can see that the stoichiometric ratio between magnesium and hydrogen is 1:1. This means that for every 1 mole of magnesium reacted, 1 mole of hydrogen gas is produced.

Therefore, the number of moles of hydrogen gas produced is also 0.097 moles.

Using the ideal gas law, we can relate the number of moles of a gas to its volume at STP:

PV = nRT

P = pressure (at STP, P = 1 atm) V = volume of gas n = number of moles R = ideal gas constant (0.0821 L·atm/(mol·K)) T = temperature (at STP, T = 273.15 K)

Rearranging the equation to solve for V:

V = (nRT) / P

Substituting the values into the equation:

V = (0.097 moles) * (0.0821 L·atm/(mol·K)) * (273.15 K) / 1 atm

V ≈ 2.26 liters

Therefore, approximately 2.26 liters of hydrogen gas will be produced at STP when 2.36 grams of magnesium reacts.