Answer:
37.8 L OF CARBON MONOXIDE IS REQUIRED TO PRODUCE 18.9 L OF NITROGEN.
Explanation:
Equation for the reaction:
2 CO + 2 NO ------> N2 + 2 CO2
2 moles of carbon monoxide reacts with 2 moles of NO to form 1 mole of nitrogen
At standard temperature and pressure, 1 mole of a gas contains 22.4 dm3 volume.
So therefore, we can say:
2 * 22.4 L of CO produces 22.4 L of N2
44.8 L of CO produces 22.4 L of N2
Since, 18.9 L of Nitrogen is produced, the volume of CO needed is:
44.8 L of CO = 22.4 L of N
x L = 18.9 L
x L = 18.9 * 44.8 / 22.4
x L = 18.9 * 2
x = 37.8 L
The volume of Carbon monoxide required to produce 18.9 L of N2 is 37.8 L
The answer to this question is A
Answer:
1.76 g is the mass of Ne is in the container.
Explanation:
We use the equation given by ideal gas which follows:
where,
P = pressure of the gas = 650 mm Hg
V = Volume of the gas = 2.50 L
T = Temperature of the gas = ![25^oC=[25+273]K=298K](https://tex.z-dn.net/?f=25%5EoC%3D%5B25%2B273%5DK%3D298K)
R = Gas constant = 
n = number of moles of Ne gas = ?
Putting values in above equation, we get:
Also, molar mass of Ne = 20.1797 g/mol
So, 
<u>1.76 g is the mass of Ne is in the container.</u>
C₃H₈ + 5 O₂ → 4 H₂O + 3 CO₂
mole ratio based on balance equation of O₂ : CO₂ i s 5 : 3
C₃H₈ + 5 O₂ → 4 H₂O + 3 CO₂
∴ if moles of CO₂ = 3 moles
then moles of O₂ = (3 moles ÷ 3) × 5
= <span>5 moles </span>
Answer:
The answer is Sodium chloride.
Na is sodium and Cl is chlorine.
