Answer:
Percent yield of reaction is<em> 150%.</em>
Explanation:
Given data:
Percent yield = ?
Actual yield of SO₃ = 586.0 g
Mass of SO₂ = 705.0 g
Mass of O₂ = 80.0 g
Solution:
Chemical equation:
2SO₂ + O₂ → 2SO₃
Number of moles of SO₂:
Number of moles = mass/ molar mass
Number of moles = 586.0 g/ 64.1 g/mol
Number of moles = 9.1 mol
Number of moles of O₂:
Number of moles = mass/ molar mass
Number of moles = 80.0 g/ 32g/mol
Number of moles = 2.5 mol
Now we will compare the mole of SO₃ with O₂ and SO₂.
SO₂ : SO₃
2 : 2
9.1 : 9.1
O₂ : SO₃
1 : 2
2.5 : 2×2.5 = 5
The number of moles of SO₃ produced by oxygen are less it will limiting reactant.
Theoretical yield of SO₃:
Mass = number of moles × molar mass
Mass = 5 mol × 80.1 g/mol
Mass = 400.5 g
Percent yield of reaction:
Percent yield = actual yield / theoretical yield × 100
Percent yield = 586.0 g/ 400.5 g× 100
Percent yield = 1.5× 100
Percent yield = 150%
<u>Answer:</u> The chemical equations are given below.
<u>Explanation:</u>
The chemical equation for the reaction of lead nitrate and sodium hydroxide follows:
By Stoichiometry of the reaction:
1 mole of aqueous solution of lead nitrate reacts with 2 moles of aqueous solution of sodium hydroxide to produce 1 mole of solid lead hydroxide and 2 moles of aqueous solution of sodium nitrate.
The chemical equation for the reaction of lead hydroxide and hydroxide ions follows:
![Pb(OH)_2(s)+2OH^-(aq.)\rightarrow [Pb(OH)_4]^{2-}(aq.)](https://tex.z-dn.net/?f=Pb%28OH%29_2%28s%29%2B2OH%5E-%28aq.%29%5Crightarrow%20%5BPb%28OH%29_4%5D%5E%7B2-%7D%28aq.%29)
By Stoichiometry of the reaction:
1 mole of lead hydroxide reacts with 2 moles of aqueous solution of hydroxide ions to produce 1 mole of aqueous solution of tetra hydroxy lead (II) complex
Hence, the chemical equations are given above.
Answer:
The mass of CO22 in total is 264 g.
The atomic mass for C is 12 g / mole.
The molar mass for CO22 is (12 + (2 × 16)) = 44 g / mole.
m C = (12 / 44) × 264 = 72 g
So, there are 72 g of C in 264 g of CO2
Explanation:
there's the answer have a good day.
Answer 0.5
when reporting calculations from sig figs, report the number of sig figs in the least accurate measurement
so
((2.0265-2.02)/2.0265) x 100.00
((2.0265/2.0265) - (2.02/2.03)) X 100.00
(1.0000 -0.995) X 100.00
0.005 X 100.00
0.5
Typically the warm front is pushed upward and the cold pushed downward.
