Answer:
1.8 g
Explanation:
Step 1: Write the balanced equation
CH₃CH₃(g) + 3.5 O₂(g) ⇒ 2 CO₂(g) + 3 H₂O(g)
Step 2: Determine the limiting reactant
The theoretical mass ratio of CH₃CH₃ to O₂ is 30.06:112.0 = 0.2684:1.
The experimental mass ratio of CH₃CH₃ to O₂ is 0.60:3.52 = 0.17:1.
Thus, the limiting reactant is CH₃CH₃
Step 3: Calculate the mass of CO₂ produced
The theoretical mass ratio of CH₃CH₃ to O₂ is 30.06:88.02.
0.60 g CH₃CH₃ × 88.02 g CO₂/30.06 g CH₃CH₃ = 1.8 g
Answer:
1) 17.5 mL
Explanation:
Hello,
In this case, the reaction between sulfuric acid and potassium hydroxide is:
In such a way, we notice a 1:2 molar ratio between the acid and the base, therefore, at the equivalence point we have:
And in terms of concentrations and volumes:
Thus, we solve for the volume of acid:
Best regards.
Volume of a rectangular block= 2.30x4.01x1.82=16.78cm3
Mass of the rectangular block= 25.71 g
Density of the block will be = 25.71/16.78=1.53 g/cm3
So the
ANSWER IS 1.53 g/cm3 density
Answer:
Mass of barium sulfate = 8.17 g
Explanation:
Given data:
Mass of sodium sulfate = 4.98 g
Mass of barium sulfate produced = ?
Solution:
Na₂SO₄ + Ba(NO₃)₂ → BaSO₄ + 2NaNO₃
Moles of sodium sulfate:
Number of moles = mass/molar mass
Number of moles =4.98 g / 142.04 g/mol
Number of moles = 0.035 mol
Now we will compare the moles pf sodium sulfate and with barium sulfate.
Na₂SO₄ : BaSO₄
1 : 1
0.035 : 0.035
Mass of barium sulfate:
Mass = number of moles × molar mass
Mass = 0.035 mol ×233.4 g/mol
Mass = 8.17 g
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