Answer:
The chemical formula is I2F9
Explanation:
Answer:
67.1%
Explanation:
Based on the chemical equation, if we determine the moles of sodium carbonate, we can find the moles of NaHCO₃ that reacted and its mass, thus:
<em>Moles Na₂CO₃ - 105.99g/mol-:</em>
6.35g * (1mol / 105.99g) = 0.0599 moles of Na₂CO₃ are produced.
As 1 mole of sodium carbonate is produced when 2 moles of NaHCO₃ reacted, moles of NaHCO₃ that reacted are:
0.0599 moles of Na₂CO₃ * (2 moles NaHCO₃ / 1 mole Na₂CO₃) = 0.1198 moles of NaHCO₃
And the mass of NaHCO₃ in the sample (Molar mass: 84g/mol):
0.1198 moles of NaHCO₃ * (84g / mol) = 10.06g of NaHCO₃ were in the original sample.
And percent of NaHCO₃ in the sample is:
10.06g NaHCO₃ / 15g Sample * 100 =
<h3>67.1%</h3>
Answer:
oxygen is limiting reactant
Explanation:
Given data:
Mass of hydrogen = 16.7 g
Mass of oxygen = 15.4 g
Limiting reactant = ?
Solution:
Chemical equation:
2H₂ + O₂ → 2H₂O
Number of moles of hydrogen:
Number of moles = mass/ molar mass
Number of moles = 16.7 g/ 2 g/mol
Number of moles = 8.35 mol
Number of moles of oxygen:
Number of moles = mass/ molar mass
Number of moles = 15.4 g/ 32 g/mol
Number of moles = 0.48 mol
Now we will compare the moles of both reactant with product,
H₂ : H₂O
2 : 2
8.35 : 8.35
O₂ : H₂O
1 : 2
0.48 : 2×0.48 = 0.96 mol
The number of moles of water produced by oxygen are less so it will limiting reactant.
I’m pretty sure it would be D.
