Answer:
The answer to your question is C₂HO₃
Explanation:
Data
Hydrogen = 3.25%
Carbon = 19.36%
Oxygen = 77.39%
Process
1.- Write the percent as grams
Hydrogen = 3.25 g
Carbon = 19.36 g
Oxygen = 77.39 g
2.- Convert the grams to moles
1 g of H ----------------- 1 mol
3,25 g of H ------------- x
x = (3.25 x 1) / 1
x = 3.25 moles
12 g of C ---------------- 1 mol
19.36 g of C ---------- x
x = (19.36 x 1) / 12
x = 1.61 moles
16g of O --------------- 1 mol
77.39 g of O --------- x
x = (77.39 x 1)/16
x = 4.83
3.- Divide by the lowest number of moles
Carbon = 3.25/1.61 = 2
Hydrogen = 1.61/1.61 = 1
Oxygen = 4.83/1.61 = 3
4.- Write the empirical formula
C₂HO₃
Answer:
Mass = 42.8g
Explanation:
4 NH 3 ( g ) + 5 O 2 ( g ) ⟶ 4 NO ( g ) + 6 H 2 O ( g )
Observe that every 4 mole of ammonia requires 5 moles of oxygen to obtain 4 moles of Nitrogen oxide and 6 moles of water.
Step 1: Determine the balanced chemical equation for the chemical reaction.
The balanced chemical equation is already given.
Step 2: Convert all given information into moles (through the use of molar mass as a conversion factor).
Ammonia = 63.4g × 1mol / 17.031 g = 3.7226mol
Oxygen = 63.4g × 1mol / 32g = 1.9813mol
Step 3: Calculate the mole ratio from the given information. Compare the calculated ratio to the actual ratio.
If all of the 1.9831 moles of oxygen were to be used up, there would need to be 1.9831 × 4 / 5 or 1.5865 moles of Ammonia. We have 3.72226 moles of ammonia - Far excess. Because there is an excess of Ammonia, the Oxygen amount is used to calculate the amount of the products in the reaction.
Step 4: Use the amount of limiting reactant to calculate the amount of H2O produced.
5 moles of O2 = 6 moles of H2O
1.9831 moles = x
x = (1.9831 * 6 ) / 5
x = 2.37972 moles
Mass of H2O = Molar mass * Molar mass
Mass = 2.7972 * 18
Mass = 42.8g
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