Answer:
= 3.78 g H₂O
Explanation:
2C₂H₆ + 3O₂ => 4CO₂ + 6H₂O
2.1g C₂H₆ = 2.1g/30.0 g/mol = 0.07 mole ethane
3.68g O₂ = 3.68g/32 g/mol = 0.115 mole oxygen
Limiting Reactant:
A quick way to determine limiting reactant is to divide moles of reactant by its respective coefficient in the balanced molecular equation. The smaller value is the limiting reactant.
moles ethane = 0.07 mole / 2 (the coefficient in balanced equation) = 0.035
moles oxygen = 0.115 mole / 3 (the coefficient in balanced equation) = 0.038
Since the smaller value is associated with ethane, then ethane is the limiting reactant and the problem is worked from the 0.07 moles of ethane in an excess of O₂.
From the equation stoichiometry ...
2 moles C₂H₆ in an excess of O₂ => 6 moles H₂O
then 0.07 mole C₂H₆ in an excess of O₂ => 6/2(0.07 moles H₂O = 0.21 mole
Converting to grams of water produced
= 0.21 mole H₂O X 18 g/mol = 3.78 g H₂O
Answer:
I think it is copper
Explanation:
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Hello! Before I answer to your question, please be sure to include a picture or else a report will be filed for your question would be commenced. You are lucky I have this for my homework tonight and I figured it out. Thank you:
The answer to your question would be as followed:
The most precise measurement for the cylinder you are mentioning is <em><u>B. 43.0mL</u></em>
Answer:
0.54 L
Explanation:
Given that,
Initial volume, V₁ = 0.5 L
Initial temperature, T₁ = 10°C = 283 K
Final temperature, T₂ = 35 C = 308 K
We need to find the final volume. The relation between the volume and temperature is given by :
So, the new volume is 0.54 L.
