Answer:
0.299 moles of PCl5
Explanation:
First form an equation based of given information: P4 + Cl2= 4PCl5
Next balance it: P4 + 10Cl2 = 4PCl5
Take your given value (53.0g of Cl2) and divide it by its molar mass to get moles of Cl2: 53.0g/70.906g = 0.747 moles Cl2
Then, multiply by the molar ratio (10Cl2 to 4PCl5): (0.747 x 4 mol PCl5)/10 mol Cl2 = 0.299 moles PCl5
Answer is: identity of the metal is gold (Au).
ω(Cl) = 35.06% ÷ 100%.
ω(Cl) = 0.3506; mass percentage of chlorine.
If we take 100 grams of the compound:
m(Cl) = ω(Cl) · m(compound).
ω(Cl) = 0.3506 · 100 g.
ω(Cl) = 35.06 g.
n(Cl) = m(Cl) ÷ M(Cl).
n(Cl) = 35.06 g ÷ 35.45 g/mol.
n(Cl) = 0.99 mol; amount of substance.
In molecule MCl₃: n(M) : n(Cl) = 1 : 3.
n(M) = 0.33 mol; amount of unknown metal.
M(M) = m(M) ÷ n(M).
M(M) = (100 g - 35.06 g) ÷ 0.33 mol.
M(M) = 196.8 g/mol; molar mass of the gold.
Easy stoichiometry conversion :)
So, for stoichiometry, we always start with our "given". In this case, it would be the 10.0 grams of NaHCO3. This unit always goes over 1.
So, our first step would look like this:
10.0
------
1
Next, we need to cancel out grams to get to moles. To do this, we will do grams of citric acid on the BOTTOM of the next step, so it cancels out. This unit in grams will be the mass of NaHCO3, which is 84.007. Then, we will do our unit of moles on top. Since this is unknown, it will be 1.
So, our 2nd step would look like this:
1 mole CO2
-----------------
84.007g NaHCO3
When we put it together: our complete stoichiometry problem would look like this:
10.0g NaHCO3 1mol CO2
---------------------- x -------------------------
1 84.007g NaHCO3
Now to find our answer, all we need to do is:
Multiply the two top numbers together (which is 10.0)
Multiply the two bottom numbers together (Which is 84.007)
And then....
Divide the top answer by the bottom answer.
10.0/84.007 is 0.119
So, from 10.0 grams of citric acid, we have 0.119 moles of CO2.
Hope I could help!
The answer is Cs and F. S and O are both non-metals and if I am not mistaken they are both on the right side of the periodic table. S and O exhibit a covalent bond, the other pairs do too except for the first pair of atoms. Therefore, the answer is A. Cs and F.
