5.74 g of mg reacts with hcl calculate the moles of h from 0.0840 and excess hcl
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5.74g of mg reacts with hcl to produce mgcl and h calculate the number of moles of h from 0.0840 mg and excess hcl
To calculate the moles of hydrogen (H₂) produced from 0.0840 moles of HCl and excess HCl, you need to consider the balanced chemical equation for the reaction between magnesium (Mg) and hydrochloric acid (HCl).
The balanced chemical equation for the reaction is:
Mg + 2HCl → MgCl₂ + H₂
From the equation, you can see that 1 mole of magnesium reacts with 2 moles of hydrochloric acid to produce 1 mole of hydrogen gas.
First, convert the mass of Mg given (5.74 g) to moles by using the molar mass of magnesium. The molar mass of Mg is 24.31 g/mol.
Moles of Mg = Mass of Mg / Molar Mass of Mg
Moles of Mg = 5.74 g / 24.31 g/mol ≈ 0.236 moles of Mg
Since magnesium is in excess, it will completely react with all the available hydrochloric acid. So, you can disregard the mass of Mg and focus on the moles of HCl provided (0.0840 moles).
Now, use the stoichiometry relationship from the balanced equation to determine the moles of hydrogen gas produced. From the balanced equation, the ratio between HCl and H₂ is 2:1.
Moles of H₂ = Moles of HCl / 2
Moles of H₂ = 0.0840 moles / 2 ≈ 0.0420 moles of H₂
Therefore, the moles of hydrogen gas produced from 0.0840 moles of HCl and excess HCl is approximately 0.0420 moles of H₂.