Answer:
Percent yield of CO₂ is 6.25 %.
Explanation:
Given data:
Percent yield of CO₂ = ?
Actual yield of CO₂ = 28.16 g
Number of moles of C₈H₁₈ = 8.000 mol
Number of moles of O₂ = 16.00 mol
Solution:
Chemical equation:
2C₈H₁₈ + 25O₂ → 16CO₂ + 18H₂O
Now we will compare the moles of C₈H₁₈ and O₂ with CO₂.
C₈H₁₈ : CO₂
2 : 16
8.000 : 16/2×8.000 = 64 mol
O₂ : CO₂
25 : 16
16 : 16/25×16= 10.24 mol
Less number of moles of CO₂ are produced from 16 moles of O₂. it will limit the yield of CO₂.
Grams of CO₂ produced:
Mass = number of moles × molar mass
Mass = 10.24 mol × 44 g/mol
Mass = 450.56 g
Percentage yield of CO₂:
Percentage yield = actual yield / theoretical yield × 100
Percentage yield = 28.16 g/ 450.56 g× 100
Percentage yield = 6.25 %
Answer:
Option E is the correct one
Explanation:
K₂Cr₂O₇ → Potassium dichromate
A ionic salt which its molar mass is 294.18 g/m
It is composed by 2 mol of K, 2 mol of Cr and 7 mol of O
2 mol of K = 78.2 g
2 mol of Cr = 104 g
7 mol of O = 112 g
In 294.2 g of compound we have __78.2 g K __ 104 g Cr ___ 112 g O
Let's calculate the percent
(78.2 / 294.2) .100 = 26.59% K
(104 / 294.2) .100 = 35.35% Cr
(112 / 294.2) . 100 = 38.06 % O
Morality= wt\m.wt*1000\v(ml)
So the m.wt of h3po4 is 98.00g\ml
Molarity = 32.7g\98.00*1000\455ml
=0.7176
You would find the highest pressure at the bottom
Molarity = moles÷liters
4.53 moles ÷ 2.85 liters = 1.59 M