Answer:
2NaOH + CO2 -> Na2CO3 + H2O
1) Find the moles of each substance
2) Determine the limitting reagent
∴ Carbon dioxide is limitting as it has a smaller value.
3) multiply the limiting reagent by the mole ratio of unknown over known
n(H2O ) = 0.3976369007 × 1/2
= 0.1988184504 moles
4) Multiply the number of moles by the molar mass of the substance.
m = 0.1988184504 × (1.008 × 2 + 16.00)
= 0.1988184504 × 18.016
= 3.581913202 g
Explanation:
Explanation:
Whenever we need to make a dilute solution of an acid then it is necessary to add water or non-acidic component into the acid first. This is because addition of water or non-acidic component directly into the acid could be highly exothermic in nature.
As a result, the acid can splutter and can cause burning of skin and other serious damage.
So, in order to avoid such type of damage the addition of water or non-acidic component into the acid actually helps to minimize the heat generated.
Thus, we can conclude that correct order of steps for making a more dilute solution of an acid is that either add all of the water or non-acid component first, or add a significant portion, before adding the acid to the mixture.
Answer : 0.8663 Kg of chalcopyrite must be mined to obtained 300 g of pure Cu.
Solution : Given,
Mass of Cu = 300 g
Molar mass of Cu = 63.546 g/mole
Molar mass of 
= 183.511 g/mole
- First we have to calculate the moles of Cu.
The moles of Cu = 4.7209 moles
From the given chemical formula, we conclude that the each mole of compound contain one mole of Cu.
So, The moles of Cu = Moles of = 4.4209 moles
- Now we have to calculate the mass of .
Mass of = Moles of × Molar mass of = 4.4209 moles × 183.511 g/mole = 866.337 g
Mass of = 866.337 g = 0.8663 Kg (1 Kg = 1000 g)
Therefore, 0.8663 Kg of chalcopyrite must be mined to obtained 300 g of pure Cu.
It's probably animal if that's what's you're asking.
