The answer is the letter C
Answer:
6.05g
Explanation:
The reaction is given as;
Ethane + oxygen --> Carbon dioxide + water
2C2H6 + 7O2 --> 4CO2 + 6H2O
From the reaction above;
2 mol of ethane reacts with 7 mol of oxygen.
To proceed, we have to obtain the limiting reagent,
2,71g of ethane;
Number of moles = Mass / molar mass = 2.71 / 30 = 0.0903 mol
3.8g of oxygen;
Number of moles = Mass / molar mass = 3.8 / 16 = 0.2375 mol
If 0.0903 moles of ethane was used, it would require;
2 = 7
0.0903 = x
x = 0.31605 mol of oxygen needed
This means that oxygen is our limiting reagent.
From the reaction,
7 mol of oxygen yields 4 mol of carbon dioxide
0.2375 yields x?
7 = 4
0.2375 = x
x = 0.1357
Mass = Number of moles * Molar mass = 0.1357 * 44 = 6.05g
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Initial concentration of magnesium nitrate M1 = 2.13 M
Initial volume of magnesium nitrate, MgNO3 V1 = 1.24 L
Final concentration of MgNO3, M2 = 1.60 M
Let the final volume of MgNO3 upon dilution be V2
Formula to use:
M1*V1 = M2*V2
V2 = M1*V1/M2
= 2.13 M * 1.24 L/1.60 M = 1.65 L
Thus, the final volume of magnesium nitrate solution upon dilution is 1.65 L
