Answer:
18.8 g
Explanation:
The equation of the reaction is;
AgClO3(aq) + LiBr(aq)------>LiClO3(aq) + AgBr(s)
Number of moles of AgClO3 = 117.63 g/191.32 g/mol = 0.6 moles
Number of moles of LiBr = 10.23 g/86.845 g/mol = 0.1 moles
Since the molar ratio is 1:1, LiBr is the limiting reactant
Molar mass of solid AgBr = 187.77 g/mol
Mass of precipitate formed = 0.1 moles * 187.77 g/mol
Mass of precipitate formed = 18.8 g
The balanced chemical reaction would be
<span>pcl3(g) + cl2(g) = pcl5(g)
First, we need to determine the theoretical yield of the reaction assuming that all of the limiting reactant is used up. Then, we divide the actual yield to the theoretical yield to determine the percent yield.
Theoretical yield = 61.3 g Cl2 ( 1 mol / 70.9 g) ( 1 mol PCl5 / 1 mol Cl2 ) ( 208.2 g / 1 mol ) = 180.01 g PCl5
Percent yield = 119.3 g / 180.01 g x100 = 66.3%</span>
The empirical formula is K₂CO₃.
The empirical formula is the <em>simplest whole-number ratio of atoms</em> in a compound.
The ratio of atoms is the same as the ratio of moles, so our job is to calculate the <em>molar ratio of K:C:O</em>.
I like to summarize the calculations in a table.
<u>Element</u> <u>Moles</u> <u>Ratio</u>¹ <u>Integers</u>²
K 0.104 2.00 2
C 0.052 1.00 1
O 0.156 3.00 3
¹ To get the molar ratio, you divide each number of moles by the smallest number.
² Round off the number in the ratio to integers to integers (2, 1, and 3).
The empirical formula is K₂CO₃.