4Al(s) + 3O2(g) --> 2Al2O3(s) This is the balanced.
From the equation:
4 moles of Al required 3 moles of O2 to produce 2 moles of Al2O3
3 moles of O2 reacted with 4 moles of Al to produce 2 moles of Al2O3
1 mole of O2 reacted with 4/3 moles of Al to produce 2/3 moles of Al2O3 (Divide by 3)
4.5 moles of O2 reacted with (4/3 *4.5) moles of Al to produce (2/3*4.5) moles of Al2O3
4.5 moles of O2 reacted with 6moles of Al to produce 3moles of Al2O3
(3) is the answer. 6 mol of Al.
So to put them all in the same units we have
<span>2500 mL </span>
<span>250 mL </span>
<span>25mL </span>
<span>2,500,000,000mL </span>
<span>So the third one is the smallest</span>
Answer:
I think it A good luck this thing is making me write 20 letters so I can give you the answer byeee
The molar mass of aluminum sulftae is 342.14 g/mol.
Since the subscript shows that there are 3 sulfurs within the substance, the total mass of sulfur is 96.21g/mol
Now take the mass of the sulfur and divide it by the molar mass of aluminum sulfate, then multiply by 100:
(96.21/342.15)(100) = 28.1% mass composition of sulfate
To solve this, let's assume ideal gas behavior.
PV=nRT
Let's solve for n. Convert units to SI units first.
Pressure = 833 torr(101325 Pa/760 torr) = 111,057.53 Pa
Volume = 250 mL(1 L/1000 mL)(1 m³/1000 L) = 2.5×10⁻⁴ m³
Temperature = 42.4 + 273 = 315.4 K
n = (8,314 J/mol·K)(315.4 K)/(111057.53 Pa)(2.5×10⁻⁴ m³)
n = 94.45 mol
The molar mass of ammonia is 17.031 g/mol.
Mass = 94.45*17.031 = <em>1,608.51 g ammonia</em>
