Answer:
Mass = 14.3 g
Explanation:
Given data:
Mass of Mg(OH)₂ = 16.0 g
Mass of HCl = 11.0 g
Mass of MgCl₂ = ?
Solution:
Chemical equation:
Mg(OH)₂ + 2HCl → MgCl₂ + 2H₂O
Number of moles of Mg(OH)₂ :
Number of moles = mass/ molar mass
Number of moles = 16.0 g/ 58.3 g/mol
Number of moles = 0.274 mol
Number of moles of HCl :
Number of moles = mass/ molar mass
Number of moles = 11.0 g/ 36.5 g/mol
Number of moles = 0.301 mol
Now we will compare the moles of Mg(OH)₂ and HCl with MgCl₂.
Mg(OH)₂ : MgCl₂
1 : 1
0.274 : 0.274
HCl : MgCl₂
2 : 1
0.301 : 1/2×0.301 = 0.150
The number of moles of MgCl₂ produced by HCl are less so it will limiting reactant.
Mass of MgCl₂:
Mass = number of moles × molar mass
Mass = 0.150 × 95 g/mol
Mass = 14.3 g
Answer:
Nickel(III) oxide is the chemical name of the compound represented by the formula Ni2O3?
double-displacement reaction
Explanation:
We have the chemical reaction:
Na₂S (aq) + Cd(NO₃)₂ (aq) → CdS (s) + 2 NaNO₃ (aq)
where:
(aq) - aqueous
(s) - solid
This is a double-displacement reaction because the reactants exchange atoms or group of atoms between themselves to form the products. To drive the reaction to the right, one of the products is a precipitate.
Generally we can express the double-displacement reaction as following:
AB + CD → AC + BD
Learn more about:
types of chemical reactions
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-Pure magnesium is commonly made by separating it from seawater. This process is known as electrolysis. The liquid magnesium formed is cooled into convenient blocks of metal known as ingots. The chlorine gas is recycled to form hydrochloric acid for the production of more magnesium chloride.
Answer: 83%
Explanation:
The detailed solution is shown in the image attached. First we must work out the balanced reaction equation because accurate solution of the problem must be based on the stoichiometry of the reaction. From the given concentration and volume of reactants, we calculate the amount of substance reacted hence identify the limiting reactant. Lastly we use simple proportion to obtain the theoretical yield of the precipitate. This is now used to calculate the actual yield as shown in the solution attached.
