How many grams of magnesium chloride should be produced by the single displacement reaction between 0.578 grams of magnesium metal and an excess hydrochloric acid?
See your other post on Al and O2.
I want to make sure I have the correct double replacement formula.
would it be
HNO3 + Al2O3 ---> Al(NO3)3 + H3O?
wrong formula, would it be...
MgCl2 ---> Mg + HCl
reverse that and balance.
Mg + 2HCl ==> MgCl2.
To determine the number of grams of magnesium chloride produced in the reaction, we need to first write a balanced chemical equation for the single displacement reaction between magnesium and hydrochloric acid.
The balanced chemical equation for this reaction is:
Mg + 2HCl -> MgCl2 + H2
From the equation, we can see that 1 mole of magnesium reacts with 2 moles of hydrochloric acid to produce 1 mole of magnesium chloride.
Now, let's calculate the number of moles of magnesium in 0.578 grams of magnesium metal. To do this, we divide the given mass by the molar mass of magnesium.
The molar mass of magnesium (Mg) is 24.31 g/mol.
Number of moles of Mg = Mass of Mg / Molar mass of Mg
= 0.578 g / 24.31 g/mol
= 0.0237 mol
According to the balanced equation, the mole ratio between magnesium and magnesium chloride is 1:1. Therefore, the number of moles of magnesium chloride produced will be the same as the number of moles of magnesium.
Number of moles of MgCl2 = 0.0237 mol
To calculate the mass of magnesium chloride produced, we multiply the number of moles by the molar mass of magnesium chloride.
The molar mass of magnesium chloride (MgCl2) is 95.21 g/mol.
Mass of MgCl2 = Number of moles of MgCl2 * Molar mass of MgCl2
= 0.0237 mol * 95.21 g/mol
= 2.255 g
Therefore, approximately 2.255 grams of magnesium chloride should be produced by the single displacement reaction between 0.578 grams of magnesium metal and an excess of hydrochloric acid.