Answer:
mh cjwe cejwh cjefblewkbfhfwouf
Explanation:
jhbckwjebfljbewfjbwefbwc
You can solve this problem through dimensional analysis.
First, find the molar mass of NaHCO3.
Na = 22.99 g
H = 1.008 g
C = 12.01 g
O (3) = 16 (3) g
Now, add them all together, you end with with the molar mass of NaHCO3.
22.99 + 1.008 + 12.01 + 16(3) = 84.008 g NaHCO3. This number means that for every mole of NaHCO3, there is 84.008 g NaHCO3. In simpler terms, 1 mole NaHCO3 = 84.008 g NaHCO3.
After finding the molar mass of sodium bicarbonate, now you can use dimensional analysis to solve for the number of moles present in 200. g of sodium bicarbonate.
Cross out the repeating units which are g NaHCO3, and the remaining unit is mole NaHCO3
200. * 1 = 200
200/ 84.008 = 2.38
Notice how there are only 3 sig figs in the answer. This is because the given problem only gave three sig figs.
Your final answer is 2.38 mol NaHCO3.
B. A chemical change has occurred and energy was given off
168.96 g of carbon dioxide (CO₂)
Explanation:
The chemical reaction representing the combustion of acetylene:
2 C₂H₂ (g) + 5 O₂ (g)→ 4 CO₂ (g) + 2 H₂O (g)
number of moles = mass / molecular weight
number of moles of acetylene (C₂H₂) = 50 / 26 = 1.92 moles
Taking in account the stoichiometry of the chemical reaction, we devise the following reasoning:
if 2 moles of acetylene (C₂H₂) produces 4 moles of carbon dioxide (CO₂)
then 1.92 moles of acetylene (C₂H₂) produces X moles of carbon dioxide (CO₂)
X = (1.92 × 4) / 2 = 3.84 moles of carbon dioxide (CO₂)
mass = number of moles × molecular weight
mass of carbon dioxide (CO₂) = 3.84 × 44 = 168.96 g
Learn more about:
combustion of hydrocarbons
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