Answer:
N2 + 3H2 ———> 2NH3
As we know 1000 grams ammonia is 58.82 moles so according to unitary method,
2 mole NH3 formed by 1 mole N2 hence 58.82 NH3 will be given by 29.41 moles N2.
No. Of moles = given mass/molar mass
Implies that
Mass of nitrogen required = 29.41*28 = 823.48 grams.
Explanation:
Answer:
THE VOLUME OF THE NITROGEN GAS AT 2.5 MOLES , 1.75 ATM AND 475 K IS 55.64 L
Explanation:
Using the ideal gas equation
PV = nRT
P = 1.75 atm
n = 2.5 moles
T = 475 K
R = 0.082 L atm/mol K
V = unknown
Substituting the variables into the equation we have:
V = nRT / P
V = 2.5 * 0.082 * 475 / 1.75
V = 97.375 / 1.75
V = 55.64 L
The volume of the 2.5 moles of nitrogen gas exerted by 1.75 atm at 475 K is 55.64 L
<h3>Answer:</h3>
162.43 g of FeCl₂
<h3>
Explanation:</h3>
Step 1: Calculate mass of Fe;
As,
Density = Mass ÷ Volume
Or,
Mass = Density × Volume
Where Volume is the volume of water displaced = 10.4 mL
Putting values,
Mass = 7.86 g.mL⁻¹ × 10.4 mL
Mass = 81.744 g of Fe
Step 2: Calculate amount of FeCl₂;
The balance chemical equation is as follow,
Fe + 2 HCl → FeCl₂ + H₂ ↑
According to this equation,
55.85 g (1 mol) Fe produced = 110.98 g (1 mol) of FeCl₂
So,
81.744 g Fe will produce = X g of FeCl₂
Solving for X,
X = (81.744 g × 110.98 g) ÷ 55.85 g
X = 162.43 g of FeCl₂
A the collisions between water particles and polllen grains
