Answer:
1019.27 g.
Explanation:
- For the balanced reaction:
<em>2Na + Cl₂ → 2NaCl,</em>
It is clear 2 moles of Na with 1 mole of Cl₂ to produce 2 moles NaCl.
- Firstly, we need to calculate the no. of moles of Cl₂ is needed to react with 57.5 mol Na:
2 moles of Na need → 1 mol of Cl₂, from the stichiometry.
57.5 moles of Na need → ??? mol of Cl₂.
<em>∴ The no. of moles of Cl₂ is needed to react with 57.5 mol Na =</em> (1 mol)(57.5 mol)/((2 mol) <em>= 28.75 mol.</em>
<em>∴ the mass of Cl₂ is needed to react with 57.5 mol Na = (no. of moles of Cl₂)(molar mass of Cl₂) =</em> (28.75 mol)(35.453 g/mol) <em>= 1019.27 g.</em>
Answer: 3.67 x 10^24
Explanation: 1mole of Cr contains 6.02x10^23 atoms
Therefore 6.098moles of Cr will contain = 6.098 x 6.02x10^23 = 3.67 x 10^24
Answer:
1.2 g
Explanation:
<u>For gallium:-</u>
Mass of gallium= 4.00 g
Molar mass gallium = 69.723 g/mol
The formula for the calculation of moles is shown below:
Thus,
<u>For arsenic:-</u>
Mass of arsenic = 5.50 g
Molar mass of arsenic = 74.9216 g/mol
The formula for the calculation of moles is shown below:
Thus,
According to the given reaction:
1 mole of gallium react with 1 mole of Arsenic
0.0574 mole of gallium react with 0.0574 mole of Arsenic
Moles of arsenic required = 0.0574 moles
Available moles of arsenic = 0.0734 moles
Limiting reagent is the one which is present in small amount. Thus, gallium is limiting reagent.
Moles left unreacted of arsenic = 0.0734 moles - 0.0574 moles = 0.016 moles
<u>Mass = Moles * Molar mass = 0.016 * 74.9216 g = 1.2 g</u>
Answer:
Leon
Explanation: all take up space and will have a mass to it.
Answer:
V = 16.3 L
Explanation:
Given data:
Volume of oxygen needed = ?
Temperature = standard = 273 K
Pressure = standard = 1 atm
Volume of sulfur trioxide = 32.6 L
Solution:
Chemical equation:
2SO₂ + O₂ → 2SO₃
Number of moles of SO₃:
PV = nRT
1 atm × 32.6 L = n×0.0821 atm.L/mol.K ×273 K
32.6 atm.L =n× 22.41 atm.L/mol
n = 32.6 atm.L / 22.41 atm.L/mol
n = 1.45 mol
now we will compare the moles of oxygen with sulfur trioxide.
SO₃ : O₂
2 : 1
1.45 : 1/2×1.45 = 0.725 mol
Volume of oxygen:
PV = nRT
1 atm × V = 0.725 mol×0.0821 atm.L/mol.K ×273 K
V = 16.25 atm.L /1 atm
V = 16.3 L