Balance the Chemical Equation. ITEM BANK: Move to Bottom 123456 C3H8 + O2 ⇒ H2O + CO2
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a. 13.25 g
b. 3 L
<h3>Further explanation</h3>Given
21 g of Sodium hydrogencarbonate
Reaction
2 NaHCO3 (s) → Na2CO3 (s) + H2O(l) + CO2(g)
Required
mass of residue
volume of CO2
Solution
mol NaHCO₃ (MW=84 g/mol) :
a. the residue = Na₂CO₃
From the equation, mol ratio NaHCO₃ : Na₂CO₃ = 2 : 1, so mol Na₂CO₃ :
mass Na₂CO₃ (MW=106 g/mol) :
b. From equation, mol CO₂ = 0.5 x mol NaHCO₃ = 0.125
RTP= 25 C, 1 atm ⇒ 1 mol =24 L
So volume CO₂ :
The number of mercury atoms required to produce 3.49 L of H2 gas according to the equation would be 1.89 x
From the balanced equation of the reaction:
2 moles of Hg are required to produce 1 mole of the H2 gas.
3.49 L of H2 gas was produced and the density of the gas is 0.0899 g/L
Recall that: density = mass/volume
Mass of the H2 gas = density x volume
= 3.49 x 0.0899
= 0.3138 g of H2 gas
Mole of 0.3138 g of H2 gas = mass/molar mass
= 0.3138/2
= 0.1569 mole
Since the ratio of Hg to H2 according to the equation is 2:1, it means that the mole of Hg is twice the mole of H2.
Thus, mole of Hg = 0.1569 x 2
= 0.3138 mole
According to Avogadro, 1 mole of a substance = 6.022×10^23
0.3138 mole would be:
0.3138 x 6.022 x 10^23
= 1.89 x
More on the number of atoms in molecules can be found here: brainly.com/question/19036641