mastering chem?

1)
N₂(g) + O₂(g) → 2NO(g)
Mole ratio O₂ : NO = 1 : 2
No. of moles of NO synthesized = 21.0 mol
No. of moles of O₂ required = (21.0 mol) × (1/2) = 10.5 mol

Consider the O₂ gas reacted:
Pressure, P = 950 mmHg = 950/760 atm
No. of moles, n = 10.5 mol
Gas constant, R = 0.08206 L atm / (mol K)
Temperature, T = (273 + 27) K = 300 K

Gas law: PV = nRT
Volume of O₂, V = nRT/P = 10.5 × 0.08206 × 300 / (950/760) L = 207 L


====
2)
3NiCl₂(aq) + 2K₃PO₄(aq) → Ni₃(PO₄)₂(s) + 6KCl(aq)
Mole ratio NiCl₂ : K₃PO₄ = 3 : 2

No. of milli-moles of NiCl₂ reacted = (0.0102 mmol/mL) × (114 mL) = 1.1628 mmol
No. of milli-moles of K₃PO₄ necessary = (1.1628 mmol) × (2/3) = 0.7752 mmol
Volume of K₃PO₄ necessary = (0.7752 mmol) / (0.205 mmol/mL) = 3.78 mL

Always begin with a balanced equation:
N2 + O2 --> 2 NO
Use the moles of NO and the coefficients of the balanced equation to calculate moles of O2 required:
21.0 mol NO X (1 mol O2 / 2 mol NO) = 10.5 mol O2
Now, use the ideal gas law to calculate the volume of O2 required:
PV = n RT
P = 950/ 760 = 1.25 atm
T = 300 K
1.25 atm (V) = 10.5 mol (0.08206 Latm/molK) (300 K)
V = 206 L O2

2) The balanced equation for this precipitation reaction is:
3 NiCl2(aq) + 2 K3PO4(aq) --> Ni3(PO4)2(s) +6 KCl(aq)

Calculate moles NiCl2:
0.114 L X 0.0102 mol/L = 1.16X10^-3 mol NiCl2

Use coefficients in the equation to calculate moles K3PO4 required:
1.16X10^-3 mol NiCl2 X (2 mol K3PO4 / 3 mol NiCl2) = 7.75X10^-4 mol K3PO4

Use the molarity to calculate volume:
7.75X10^-4 mol X (1 L/0.205 mol) = 3.78X10^-3 L = 3.78 mL K3PO4