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>
1 significant figure, because there is no decimal after the zero the zero doesn't count.
Group 17
Please brainliesttt
6g of hydrogen gas is my answer. I'm sorry if I'm wrong.
The solution is 45.7 % (NH₄)₂CO₃ by mass.
Mass of solution = 0.0332 kg + 0.0395 kg = 0.0727 kg
% (NH₄)₂CO₃ = Mass of (NH₄)₂CO₃/Total mass × 100 % = 0.0332 kg/0.0727 kg × 100 % = 45.7 %
