Answer:
441.28 g Oxygen
Explanation:
- The combustion of hydrogen gives water as the product.
- The equation for the reaction is;
2H₂(g) + O₂(g) → 2H₂O(l)
Mass of hydrogen = 55.6 g
Number of moles of hydrogen
Moles = Mass/Molar mass
= 55.6 g ÷ 2.016 g/mol
= 27.8 moles
The mole ratio of Hydrogen to Oxygen is 2:1
Therefore;
Number of moles of oxygen = 27.5794 moles ÷ 2
= 13.790 moles
Mass of oxygen gas will therefore be;
Mass = Number of moles × Molar mass
Molar mass of oxygen gas is 32 g/mol
Mass = 13.790 moles × 32 g/mol
<h3> = 441.28 g</h3><h3>Alternatively:</h3>
Mass of hydrogen + mass of oxygen = Mass of water
Therefore;
Mass of oxygen = Mass of water - mass of hydrogen
= 497 g - 55.6 g
<h3> = 441.4 g </h3>
Answer:
0.508 L of solution.
Explanation:
Always a safe bet to convert to moles:

Where n is moles, m is mass, and MM is molar mass.
Now remember the equation for concentration (molarity):

Where C is the concentration, n is moles, and V is volume.
To make this easy, combine the two equations (note n appears in both, so you can do a substitution) and solve for V as the question asks:

Therefore we can make 0.508 L of solution.
Answer:
If 13.4 grams of nitrogen gas reacts we'll produce 16.3 grams of ammonia
Explanation:
Step 1: Data given
Mass of nitrogen gas (N2) = 13.4 grams
Molar mass of N2 = 28 g/mol
Molar mass of NH3 = 17.03 g/mol
Step 2: The balanced equation
N2 + 3H2 → 2NH3
Step 3: Calculate moles of N2
Moles N2 = Mass N2 / molar mass N2
Moles N2 = 13.4 grams / 28.00 g/mol
Moles N2 = 0.479 moles
Step 4: Calculate moles of NH3
For 1 mol N2 we need 3 moles H2 to produce 2 moles NH3
For 0.479 moles N2 we'll produce 2*0.479 = 0.958 moles
Step 5: Calculate mass of NH3
Mass of NH3 = moles NH3 * molar mass NH3
Mass NH3 = 0.958 moles * 17.03 g/mol
Mass NH3 = 16.3 grams
If 13.4 grams of nitrogen gas reacts we'll produce 16.3 grams of ammonia