Answer: 6.78 moles
Explanation:
Assuming the actual question is about K2CO3,
mw 99.1
672/99.1 moles
Answer:
70.34 litres
Explanation:
Using ideal gas law equation;
PV = nRT
Where;
P = pressure (atm)
V = volume (L)
n = number of moles (mol)
R = gas law constant (0.0821 Latm/molK)
T = temperature (K)
According to the information given in the question;
n = 3 moles
V = ?
P = 105.2 KPa = 105.2/101 = 1.04atm
T = 24.0°C = 24 + 273 = 297K
Using PV = nRT
1.04 × V = 3 × 0.0821 × 297
1.04V = 73.15
V = 73.15 ÷ 1.04
V = 70.34 litres.
Answer:
39.72 g
Explanation:
Given data:
Mass of CsF = 15.2 g
Mass of XeF₆ = 260 g
Mass of Cs[XeF₇] = ?
Solution:
Chemical reaction:
CsF + XeF₆ → Cs[XeF₇]
Number of moles of CsF:
Number of moles = mass/ molar mass
Number of moles = 15.2 g/151.9 g/mol
Number of moles = 0.1 mol
Number of moles of XeF₆ :
Number of moles = mass/ molar mass
Number of moles = 260 g/245.28 g/mol
Number of moles = 1.06 mol
Now we will compare the moles of Cs[XeF₇] with both reactants.
CsF : Cs[XeF₇]
1 : 1
0.1 : 0.1
XeF₆ : Cs[XeF₇]
1 : 1
1.06 ; 1.06
Number of moles of Cs[XeF₇] produce by CsF are less so it will limiting reactant and limit the yield of Cs[XeF₇].
Mass of Cs[XeF₇]:
Mass = number of moles × molar mass
Mass = 0.1 mol × 397.2 g/mol
Mass = 39.72 g