Answer:
Mass = 20,000 g
Explanation:
Given data:
Mass of MgO formed = ?
Mass of Mg react = 12 Kg (12 Kg × 1000/1 Kg = 12000 g)
Solution:
Chemical equation:
2Mg + O₂ → 2MgO
Number of moles of Mg:
Number of moles = mass/molar mass
Number of moles = 12000 g/ 24 g/mol
Number of moles = 500 mol
Now we will compare the moles of Mg and MgO.
Mg : MgO
2 : 2
500 : 500
Mass of MgO:
Mass = number of moles × molar mass
Mass = 500 mol × 40 g/mol
Mass = 20,000 g
Molar mass of N2H4 = 32 grams/mole
<span>3.95 grams of N2H4 = 3.95/32
= 0.123 moles </span>
<span>This will produce 0.123 moles of N2 </span>
<span>Now,
From the gas law equation. </span>
<span>P.V = n x R x T </span>
<span>P = 1 atm (given)
V = </span><span>0.123</span><span> x 0.082057 x 295 </span>
<span>V = 2.97 Liters </span>
<span>Theoretical yield = 2.97 Liters.
Actual yield = 0.750 Liters </span>
percentage yield = (0.75/2.97) x 100 %
= 25.25 %
Erosion or Weathering. Probably erosion.
Ok,
So what don't you understand. The whole page
Answer:
Na3PO4 is excess reactant, CaCl2 is limiting reactant.
Explanation:
3CaCl2 + 2Na3PO4 ---> Ca3(PO4)2 + 6NaCl
from reaction : 3 mol 2 mol
given: 6 mol 5 mol (X)
X = (6*2)/3 = 4 mol Na3PO4
For 6 mol CaCl2 we need 4 mol Na3PO4, but we have 5 mol Na3PO4,
Na3PO4 is excess reactant, so CaCl2 is limiting reactant.
