Hydrated salts are when salt crystals have water molecules bound. Anhydrous salts are when the water has been removed.
mass of water removed = hydrated salt - anhydrate salt
= 11.75 g - 9.25 g = 2.50 g
number of water moles = 2.50 g / 18 g/mol = 0.139 mol
number of cobalt (II) chloride moles = 9.25 g / 130 g/mol = 0.0712 mol
ratio of water moles to CoCl₂ moles - 0.139 mol / 0.0712 mol = 1.95
rounded off 2 moles of water for every 1 mol of CoCl₂
formula - CoCl₂.2H₂O
name - Cobalt(II) chloride dihydrate
First figure out how many grams must freeze and then convert the grams to moles.
<span>Hf = -334 J/g. Convert this to KJ/g by dividing by 1000. (There are 1000 Joules in a kJ). </span>
<span>Hf = -334 J/g ÷ 1000 J/kj = -0.334 kJ/g </span>
<span>Now, divide 100 kJ by -0.334 kJ/g (see how the units are lining up?) </span>
<span>100 kJ ÷ -0.334 kJ/g = 299 g </span>
<span>Now convert this to moles by dividing by the molecular weight of water (18.0g/mole). </span>
<span>299 ÷ 18.0 = 16.6 moles </span>
C, to make sure the design works as expected.
A prototype is first, typical model of the said product. Hope this helps!
Answer:
Sr would be the limiting reactant
5 moles
Explanation:
Since the equation is a balanced equation, the coefficient shows how each substance relates to the other in terms of the number of moles.
Reactants would be those on the left hand side of the arrow, while the products would be found on te right and side of the arrow. In this question, the reactants would be Sr and O₂.
Limiting reactant is the reactant that is insufficient; meaning to say that there is not enough of that substance and thus the reaction cannot continue. The other reactant(s) that is not limiting is called the excess reactants.
From the balanced equation, 2 moles of Sr is needed to react with 1 mole of O₂. Thus, if we have 5 moles of each reactant, Sr would be the limiting reactant since for every 1 mole of O₂, there has to be 2 moles of Sr in order for the reaction to proceed. Thus, if we have 5 moles of O₂, we would need 10 moles of Sr.
When we work out the amount of products formed, we look at the number of moles of the limiting reactant. This is because the limiting reactant determines how much is being reacted, while the excess number of moles of the excess reactant will remain unreacted.
For every 2 moles of Sr reacted, 2 moles of SrO would be produced. This means that the mole ratio of Sr to SrO is 1:1. Thus, since 5 moles of Sr has been reacted, 5 moles of the product (SrO) would be produced.