<span>4 Na + O</span>₂<span> = 2 Na</span>₂<span>O
</span>
4* 23 g Na --------> 16 g O₂
9.5 g Na ------------> ?
Mass of O₂ = 9.5 * 16 / 4 * 23
Mass = 152 / 92
Mass = 1.6521 g of O₂
Molar mass O₂ = 16.0 g/mol
1 mole O₂ ------------ 16.0 g
? mole O₂ ------------ 1.6521 g
mole O₂ = 1.6521 * 1 / 16.0
≈ 0.10325 moles of O₂
hope this helps!
Ti + 2 Cl2 → TiCl4
(3.00 g Ti) / (47.867 g Ti/mol) = 0.062674 mol Ti
(6.00 g Cl2) / (70.9064 g Cl2/mol) = 0.084619 mol Cl2
0.084619 mole of Cl2 would react completely with 0.084619 x (1/2) = 0.0423095 mole of Ti, but there is more Ti present than that, so Ti is in excess and Cl2 is the limiting reactant.
(0.084619 mol Cl2) x (1 mol TiCl4 / 2 mol Cl2) x (189.679 g TiCl4/mol) = 8.025 g TiCl4 in theory
(7.7 g) / (8.025 g) = 0.96 = 96% yield TiCl4
Answer:
an endothermic reaction
Explanation:
An endothermic reaction has occurred.
The epicenter was located somewhere on a circle centered at Recording station X, with a radius of 250 km.<span>
</span>
At STP, the volume of a gas represents the number of particles.That said, from the chemical reaction one mole of oxygen reacts with two moles of co to produce the product, CO2At STP, 3 moles of Oxygen will produce 6 moles of CO2. Hence It follows that at standard temperature and pressure, 6.0 L of CO2 will be produced. Option D.
