Answer:
Hydroxylamine and pyridine
Explanation:
Just did it
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
Answer:
B
Explanation:
Molarity = 0.010M
Volume = 2.5L
Applying mole-concept,
0.010mole = 1L
X mole = 2.5L
X = (0.010 × 2.5) / 1
X = 0.025moles
0.025moles is present in 2.5L of NaOH solution.
Molar mass of NaOH = (23 + 16 + 1) = 40g/mol
Number of moles = mass / molar mass
Mass = number of moles × molar mass
Mass = 0.025 × 40
Mass = 1g
1g is present in 2.5L of NaOH solution
