Answer:
704.6 g CO2
Explanation:
MM sucrose = 342.3 g/mol
MM CO2 = 44.01 g/mol
g CO2 = 456.7 g sucrose x (1 mol sucrose/MM sucrose) x (12 moles CO2/1 mol sucrose) x (MM CO2/1mol CO2) = 704.6 g CO2
Answer:
55.75g
Explanation:
From
m/M = CV
Where
m= required mass of solute
M= molar mass of solute
C= concentration of solution
V= volume of solution=675ml
Molar mass of solute= 3(23) + 31 + 4(16)= 69+31+64=164gmol-1
Number of moles of sodium ions present= 1.5× 675/1000= 1.01 moles
Since 1 mole of Na3PO4 contains 3 moles of Na+
It implies that 1.01/3 moles of Na3PO4 are present in solution= 0.34moles
mass of Na3PO4= number of moles × molar mass= 0.34 × 164 =55.75g
Moles of Oxygen= 2.8075 moles
<h3>Further explanation</h3>
Given
29.2 grams of acetylene
Required
moles of Oxygen
Solution
Reaction(Combustion of Acetylene) :
2 C₂H₂ (g) + 5 O₂ (g) ⇒ 4CO₂ (g) + 2H₂O (g)
Mol of Acetylene :
= mass : MW Acetylene
= 29.2 g : 26 g/mol
= 1.123
From equation, mol ratio of Acetylene(C₂H₂) : O₂ = 2 : 5, so mol O₂ :
= 5/2 x mol C₂H₂
= 5/2 x 1.123
= 2.8075 moles
F. hold on to their protons more strongly
