The reaction equation is:
CaF₂ + H₂SO₄ → 2HF + CaSO₄
The molar ratio between fluorite and hydrogen fluoride is 1 : 1.
The moles of fluorite supplied are:
Moles = 15.6 / 78.07
Moles = 0.200
The moles of hydrogen fluoride produced will be 0.2.
Now, we may use the ideal gas equation to determine the temperature:
PV = nRT
T = PV/nR
T = (899 * 7.4) / (0.2 * 62.36)
T = 533.40 K
The temperature will be 260.25 °C
find mol of N2 present using gas law equation
PV = nRT
P = pressure = 688/760 = 0.905 atm.
V = 100mL = 0.1L
n = ???
R = 0.082057
T = 565+273 = 838
Substitute:
0.905*0.1 = n*0.082057*838
n = 0.0905 / 68.76
n = 0.00132 mol N2
Molar mass N2 = 28 g/mol
0.00132 mol = 0.00132*28 = 0.037g N2 gas
Answer:
0.4694 moles of CrCl₃
Explanation:
The balanced equation is:
Cr₂O₃(s) + 3CCl₄(l) → 2CrCl₃(s) + 3COCl₂(aq)
The stoichiometry of the equation is how much moles of the substances must react to form the products, and it's represented by the coefficients of the balanced equation. So, 1 mol of Cr₂O₃ must react with 3 moles of CCl₄ to form 2 moles of CrCl₃ and 3 moles of COCl₂.
The stoichiometry calculus must be on a moles basis. The compounds of interest are Cr₂O₃ and CrCl₃. The molar masses of the elements are:
MCr = 52 g/mol
MCl = 35.5 g/mol
MO = 16 g/mol
So, the molar mass of the Cr₂O₃ is = 2x52 + 3x35.5 = 210.5 g/mol.
The number of moles is the mass divided by the molar mass, so:
n = 49.4/210.5 = 0.2347 mol of Cr₂O₃.
For the stoichiometry:
1 mol of Cr₂O₃ ------------------- 2 moles of CrCl₃
0.2347 mol of Cr₂O₃----------- x
By a simple direct three rule:
x = 0.4694 moles of CrCl₃
<span>The mass of an object is measured in either grams or kilograms. Mass is best described as the amount of matter, or "stuff," in a solid, and is different from weight (which is the force of gravity on an object). Since mass is used with solids, it will be measured in grams or kilograms (rather than in something like liters, which would be used with the volume of a liquid). To measure mass, you can use a balance, for example a triple balance beam.</span>
