puzzling moles q :S
The equation for the reaction between magnesium hydroxide and hydrochloric acid is
shown below.
Mg(OH)2(s) + 2HCl(aq) → MgCl2(aq) + 2H2O(l)
Calculate the volume, in cm3, of 1.00 moldm–3 hydrochloric acid required to react
completely with 1.00 g of magnesium hydroxide.
the answer reads:
Mr (Mg(NO3)2 = 58(.3) where does this come from!? NO3?? there is no nitrogen in the q....
moles Mg(OH)2 = 0.0172
moles HCl = 2 × 0.0172 = 0.0344 or 0.0343 (mol)
vol HCl = 0.0343 × 1000 = 34.3 – 34.5 (cm3)
shown below.
Mg(OH)2(s) + 2HCl(aq) → MgCl2(aq) + 2H2O(l)
Calculate the volume, in cm3, of 1.00 moldm–3 hydrochloric acid required to react
completely with 1.00 g of magnesium hydroxide.
the answer reads:
Mr (Mg(NO3)2 = 58(.3) where does this come from!? NO3?? there is no nitrogen in the q....
moles Mg(OH)2 = 0.0172
moles HCl = 2 × 0.0172 = 0.0344 or 0.0343 (mol)
vol HCl = 0.0343 × 1000 = 34.3 – 34.5 (cm3)
Yeah, it's just an error...ignore it 
Original post by hamish123
The equation for the reaction between magnesium hydroxide and hydrochloric acid is
shown below.
Mg(OH)2(s) + 2HCl(aq) → MgCl2(aq) + 2H2O(l)
Calculate the volume, in cm3, of 1.00 moldm–3 hydrochloric acid required to react
completely with 1.00 g of magnesium hydroxide.
the answer reads:
Mr (Mg(NO3)2 = 58(.3) where does this come from!? NO3?? there is no nitrogen in the q....
moles Mg(OH)2 = 0.0172
moles HCl = 2 × 0.0172 = 0.0344 or 0.0343 (mol)
vol HCl = 0.0343 × 1000 = 34.3 – 34.5 (cm3)
shown below.
Mg(OH)2(s) + 2HCl(aq) → MgCl2(aq) + 2H2O(l)
Calculate the volume, in cm3, of 1.00 moldm–3 hydrochloric acid required to react
completely with 1.00 g of magnesium hydroxide.
the answer reads:
Mr (Mg(NO3)2 = 58(.3) where does this come from!? NO3?? there is no nitrogen in the q....
moles Mg(OH)2 = 0.0172
moles HCl = 2 × 0.0172 = 0.0344 or 0.0343 (mol)
vol HCl = 0.0343 × 1000 = 34.3 – 34.5 (cm3)
They meant Mg(OH)2. If you see the Mr of 58.3 is exactly the Mr for this compound
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