Answer:
Mass of C₂H₄N₂ produced = 3.64 g
Explanation:
The balanced chemical equation for the reaction is given below:
3CH₄ (g) + 5CO₂ (g) + 8NH₃ (g) → 4C₂H₄N₂ (g) + 10H₂O (g)
From the equation, 3 moles of CH₄ reacts with 5 moles of CO₂ and 8 moles of NH₃ to produce 4 moles of C₂H₄N₂ and 10 moles of H₂O
Molar masses of the compounds are given below below:
CH₄ = 16 g/mol; CO₂ = 44 g/mol; NH3 = 17 g/mol; C₂H₄N₂ = 56 g/mol; H₂O g/mol
Comparing the mole ratios of the reacting masses;
CH₄ = 1.65/16 = 0.103
CO₂ = 13.5/44 = 0.307
NH₃ = 2.21/17 = 0.130
converting to whole number ratios by dividing with the smallest ratio
CH₄ = 0.103/0.103 = 1
CO₂ = 0.307/0.103 = 3
NH₃ = 0.130/0.103 = 1.3
Multiplying through with 5
CH₄ = 1 × 5 = 5
CO₂ = 3 × 5 = 15
NH₃ = 1.3 × 5 = 6.5
Therefore, the limiting reactant is NH₃
8 × 17 g (136 g) of NH₃ reacts to produce 4 × 56 g (224 g) of C₂H₄N₂
Therefore, 2.21 g of NH₃ will produce (2.21 × 224)/136 g of C₂H₄N₂ = 3.64 g of C₂H₄N₂
Mass of C₂H₄N₂ produced = 3.64 g
Answer:
B. 1.65 L
Explanation:
Step 1: Write the balanced equation
2 SO₂(g) + O₂(g) ⇒ 2 SO₃(g)
Step 2: Calculate the moles of SO₂
The pressure of the gas is 1.20 atm and the temperature 25 °C (298 K). We can calculate the moles using the ideal gas equation.
P × V = n × R × T
n = P × V / R × T
n = 1.20 atm × 1.50 L / (0.0821 atm.L/mol.K) × 298 K = 0.0736 mol
Step 3: Calculate the moles of SO₃ produced
0.0736 mol SO₂ × 2 mol SO₃/2 mol SO₂ = 0.0736 mol SO₃
Step 4: Calculate the volume occupied by 0.0736 moles of SO₃ at STP
At STP, 1 mole of an ideal gas occupies 22.4 L.
0.0736 mol × 22.4 L/1 mol = 1.65 L
Answer:
Pollution, non native species being introduced, and other human activities
Explanation:
Human activities affect marine ecosystems as a result of pollution, overfishing, the introduction of invasive species,and acidification, which all impact on the marine food web and may lead to largely unknown consequences for the biodiversity and survival of marine life forms.
The volume of oxygen at STP required would be 252.0 mL.
<h3>Stoichiometic problem</h3>
The equation for the complete combustion of C2H2 is as below:
The mole ratio of C2H2 to O2 is 2:5.
1 mole of a gas at STP is 22.4 L.
At STP, 100.50 mL of C2H2 will be:
100.50 x 1/22400 = 0.0045 mole
Equivalent mole of O2 according to the balanced equation = 5/2 x 0.0045 = 0.01125 moles
0.01125 moles of O2 at STP = 0.01125 x 22400 = 252.0 mL
Thus, 252.0 mL of O2 gas will be required at STP.
More on stoichiometric problems can be found here: brainly.com/question/14465605
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