A sample of 5.50 g of Mg(OH)2 is added to 25.4 mL of 0.150 M HNO3. 1)Write the chemical equation for the reaction that occurs.?

a)
Mg(OH)₂ + 2HNO₃ → Mg(NO₃)₂ + 2H₂O


b)
Molar mass of Mg(OH)₂ = (24.3 + 16.0×2 + 1.0×2) g/mol = 58.3 g/mol
Initial no. of moles of Mg(OH)₂ = (5.50 g) / (58.3 g/mol) = 0.0943 mol
Initial no. of moles of HNO₃ = (0.150 mol/L) × (25.4/1000 L) = 0.00381 mol

According to the equation in a), mole ratio Mg(OH)₂ : HNO₃ = 1 : 2
If HNO₃ completely reacts, Mg(OH)₂ needed = (0.00381 mol) × (1/2) = 0.00191 mol < 0.0943 mol
Hence, Mg(OH)₂ is in excess, and HNO₃ is the limiting reactant.


c)
According to b), after the reaction is complete :
No. of moles of Mg(OH)₂ present = (0.0943 - 0.00191) mol = 0.0924 mol (to 3 sig. fig.)


d)
According to b), HNO₃ is the limiting reactant.
No. of moles of HNO₃ present after the reaction is complete = 0 mol


e)
According to the equation in a), mole ratio HNO₃ : Mg(NO₃)₂ = 2 : 1
No. of moles of HNO₃ reacted = 0.00381 mol
No. of moles of Mg(NO₃)₂ produced = (0.00381 mol) × (1/2) = 0.00191 mol

Molar mass of Mg(NO₃)₂ = (24.3 + 14.0×2 + 16.0×6) g/mol = 148.3 g/mol
Mass of Mg(NO₃)₂ produced = (0.00191 mol) × (148.3 g/mol) = 0.283 g

1)
Mg(OH)2 + 2 HNO3 → Mg(NO3)2 + 2 H2O

2)
(5.50 g Mg(OH)2) / (58.3197 g Mg(OH)2/mol) = 0.094308 mol Mg(OH)2
(0.0254 L) x (0.150 mol/L HNO3) = 0.00381 mol HNO3

0.00381 mole of HNO3 would react completely with 0.00381 x (1/2) = 0.001905 mole of Mg(OH)2, but there is more Mg(OH)2 present than that, so Mg(OH)2 is in excess and HNO3 is the limiting reactant.

3)
(0.094308 mol Mg(OH)2 initially) - (0.001905 mol Mg(OH)2 reacted) = 0.0924 mol Mg(OH)2 left over

4) No HNO3 is present after the reaction is complete, because it is the limiting reactant.

5)
(0.00381 mol HNO3) x (1 mol Mg(NO3)2 / 2 mol HNO3) = 0.001905 = 0.00191 mol Mg(NO3)2

1. Mg(OH)2) + 2 HNO3 --> Mg(NO3)2 + 2 H2O
2. Moles Mg(OH)2 = 5.50 g / 58.32 g/mol = 0.0943 mol Mg(OH)2
Moles HNO3 = 0.0254 L X 0.150 mol/L = 3.81X10^-3 mol HNO3
HNO3 will be the limiting reactant. You can see from the balanced equation that you need 2 moles of HNO3 to react with 1 mol Mg(OH)2. Because you have fewer moles of HNO3 than Mg(OH)2, HNO3 must be the limiting reactant.

3. Moles Mg(OH)2 consumed = 3.81X10^-3 mol HNO3 X (1 mol Mg(OH)2 / 2 mol HNO3) = 1.91X10^-3 mol Mg(OH)2 consumed.
Moles Mg(OH)2 remaining = 0.0943 - 1.9X10^-3 = 0.0924 mol Mg(OH)2

4. 0 moles HNO3 remain--they are all consumed.

5. You calculated in 3. that 1.91X10^-3 moles of Mg(OH)2 are consumed. From the balanced equation, there is a 1:1 relationship between moles Mg(OH)2 consumed and moles Mg(NO3) formed. Therefore,
Moles Mg(NO3)2 formed = 1.91X10^-3 moles.