In the reaction# Mg (s) + 2HCl (aq) -> H_2 (g) + MgCl_2 (aq)#, how many grams of hydrogen gas will be produced from 125.0 milliliters of a 6.0 M HCI in an excess of Mg?

To determine the grams of hydrogen gas produced, you need to use stoichiometry. First, calculate the moles of HCl using the given volume and molarity. Then, use the mole ratio from the balanced equation to find the moles of hydrogen gas. Finally, convert moles of hydrogen gas to grams using its molar mass. The molar mass of hydrogen gas (H₂) is approximately 2.016 grams/mol.

1. Calculate moles of HCl: Moles of HCl = volume (in liters) × molarity

2. Use the mole ratio from the balanced equation to find moles of hydrogen gas.

3. Convert moles of hydrogen gas to grams using its molar mass:

   Grams of hydrogen gas = moles of hydrogen gas × molar mass of hydrogen gas

To solve this problem, we can use stoichiometry to find the amount of hydrogen gas produced. First, we need to find the number of moles of HCl using its molarity and volume. Then, we use the stoichiometric coefficients from the balanced equation to find the number of moles of hydrogen gas produced. Finally, we convert the moles of hydrogen gas to grams using its molar mass.

1. Calculate the moles of HCl: [ \text{moles of HCl} = \text{Molarity} \times \text{Volume (in liters)} ]

2. Use the stoichiometric coefficients from the balanced equation to find the moles of hydrogen gas produced. [ \text{Moles of H}_2 = \frac{\text{moles of HCl}}{2} ]

3. Convert moles of hydrogen gas to grams using its molar mass: [ \text{Grams of H}_2 = \text{moles of H}_2 \times \text{molar mass of H}_2 ]

Given:

• Volume of HCl solution: 125.0 milliliters (which is 0.125 liters)
• Molarity of HCl: 6.0 M
• Molar mass of hydrogen gas ((H_2)): 2.016 g/mol

Now, let's plug the values into the equations:

1. ( \text{moles of HCl} = 6.0 , \text{M} \times 0.125 , \text{L} )
2. ( \text{moles of H}_2 = \frac{\text{moles of HCl}}{2} )
3. ( \text{Grams of H}_2 = \text{moles of H}_2 \times 2.016 , \text{g/mol} )

Calculate:

1. ( \text{moles of HCl} = 6.0 , \text{M} \times 0.125 , \text{L} = 0.75 , \text{moles} )
2. ( \text{moles of H}_2 = \frac{0.75 , \text{moles}}{2} = 0.375 , \text{moles} )
3. ( \text{Grams of H}_2 = 0.375 , \text{moles} \times 2.016 , \text{g/mol} = 0.756 , \text{grams} )

Therefore, 0.756 grams of hydrogen gas will be produced from 125.0 milliliters of a 6.0 M HCl solution in excess of magnesium.

To find the grams of hydrogen gas produced, we first need to determine the number of moles of hydrogen gas produced using the stoichiometry of the reaction. Then, we can use the molar mass of hydrogen gas to convert moles to grams. Since the reaction ratio between magnesium and hydrogen chloride is 1:2, we'll use the given concentration of HCl and the volume to find the moles of HCl reacted. Then, we'll use the stoichiometric ratio to find the moles of hydrogen gas produced. Finally, we'll convert moles of hydrogen gas to grams using its molar mass.

1. Calculate moles of HCl: ( \text{Molarity} = \frac{\text{moles}}{\text{volume (L)}} ) ( \text{moles} = \text{Molarity} \times \text{volume (L)} ) Convert milliliters to liters: ( 125.0 , \text{milliliters} = 0.125 , \text{liters} ) ( \text{moles HCl} = 6.0 , \text{M} \times 0.125 , \text{L} )

2. Use stoichiometry to find moles of ( H_2 ): From the balanced equation, 1 mole of magnesium reacts with 2 moles of HCl to produce 1 mole of ( H_2 ). So, moles of ( H_2 = \frac{\text{moles HCl}}{2} )

3. Calculate grams of ( H_2 ): ( \text{moles} = \frac{\text{grams}}{\text{molar mass}} ) Rearrange to find grams: ( \text{grams} = \text{moles} \times \text{molar mass} ) Molar mass of ( H_2 ) is approximately ( 2.016 , \text{g/mol} ) ( \text{grams of} , H_2 = \text{moles} \times 2.016 , \text{g/mol} )

Perform the calculations to find the grams of ( H_2 ).