Answer:
Hydrogen Gas
Explanation:
The hydrogen gas now in the test tube is flammable. So when I hold the tube over the flame, the hydrogen gas ignites and creates a popping sound. It is like a tiny bomb or fireworks.
Hope this helped! :)
The mass of sulfur in a sample of CaSO4 with a mass of 65.8 g is 15.50g.
<h3>How to calculate mass of an element in a compound?</h3>
According to this question, a 10.4 g sample of CaSO4 is found to contain 3.06 g of Ca and 4.89 g of O.
This means that the mass of sulfur in the 10.4g of CaSO4 is 10.4g - (3.06g + 4.89g) = 10.4g - 7.95g = 2.45g
Next, we calculate the percent ratio of each element in the compound; CaSO4.
- Ca = 3.06g/10.4g × 100 = 29.42%
- S = 2.45g/10.4g × 100 = 23.56%
- O = 4.89g/10.4g × 100 = 47.02%
According to this question, a sample of CaSO4 with a mass of 65.8 g is given. The mass of each element in this compound is as follows:
- Ca = 29.42/100 × 65.8g = 19.36g
- S = 23.56/100 × 65.8g = 15.50g
- O = 47.02/100 × 65.8g = 30.94g
Therefore, the mass of sulfur in a sample of CaSO4 with a mass of 65.8 g is 15.50g.
Learn more about mass at: brainly.com/question/13672279
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Answer:
- <em>You could expect 3.48 grams of C₂H₄N₂</em>
Explanation:
You must start by stating the chemical equation for the reaction of ammonia, carbon dioxide, and methane to produce aminoaceto nitrile.
1. Word equation:
Ammonia + Carbon dioxide + Methane → Aminoacetonitrile + Water
2. Balanced chemical equation:

3. Convert the mass of each reactant into number of moles:
<u>Formula:</u>
- Number of moles = mass in grams/molar mass
<u>2.11g NH₃</u>
- Number of moles = 2.11g / 17.03g/mol = 0.124 mol NH₃
<u>14.9g CO₂</u>
- Number of moles = 14.9g/44.01g/mol = 0.339 mol CO₂
<u>1.75g CH₄</u>
- Number of moles = 1.75g/16.04g/mol = 0.109 mol CH₄
4. Theoretical mol ratio
From the balanced chemical equation, using the coefficientes:

5. Limiting reagent
The available amounts of the reactants are:
Fom the theoretical mole ration, to react with 0.124 mol of NH₃ you would need:
- 0.124molNH₃ × (5molCO₂/8molNH₃) = 0.0775 mol CO₂
Since there are 0.339 moles available, this is in excess.
- 0.124molNH₃ × (3molCH₄/8molNH₃) = 0.0465mol CO₂
Since there are 0.109 moles available, this is in excess too.
Hence, the limiting reagent is NH₃.
6. Yield
Use the theoretical ratio:
- 0.124molNH₃ × (4molC₂H₄N₂ / 8molNH₃) = 0.0620 mol C₂H₄N₂
Convert to grams:
- Mass = number of moles × molar mass
- 0..0620 mol × 56.068g/mol = 3.48 g of C₂H₄N₂ ← answer
2 hydrogen atoms hope this helps