Download Gas Stoichiometry

Determine volume ratios for gaseous
reactants and products by using coefficients
from chemical equations.
Apply gas laws to calculate amounts of
gaseous reactants and products in a
chemical reaction.
• When gases react, the coefficients in the balanced chemical equation
represent both molar amounts and relative volumes.
 Avogadro’s Law: Equal volumes of gases contain equal numbers
of moles.
 This means that one mole of any gas will contain the same
number of moles as one mole of any other gas.
 Because of this, the coefficients in balanced equations not only
represent mole ratios, but also volume ratios of gases (and
ONLY gases.)
• Example:
2H2(g)
2 mol H2 gas
2 L H2 gas
→
2H2O(g)
means
1 mol O2 g
2 mol water vapor
and also
1 L O2 gas
2 L water vapor
+
O2(g)
• How many liters of hydrogen gas are required to react with
1.50 L oxygen gas?
•
2H2(g)
+
→
O2(g)
1.50 𝐿 𝑂2 x
2𝐿 𝐻2
1𝐿 𝑂2
2H2O(g)
= 3.00 𝐿 𝑂2
• You can use the Ideal Gas Law to derive that at STP, 1 mol of
any gas has a volume of 22.4 L. In fact, give it a shot:
• If P = 1 atm, T = 273 K, and n = 1 mol, what is V?
• PV = nRT
• (1 atm)(273 K) = (1 mol)(0.0821
• V = 22.4 L
𝐿 𝑎𝑡𝑚
)V
𝑚𝑜𝑙 𝐾
• So for gases (and gases only!), the following conversion factor
can be used:
1 mol = 22.4 L
N2 (g) + 3 H2 (g) → 2 NH3 (g)
• Find the mass of ammonia produced if 5.00 L of nitrogen gas
reacts with hydrogen gas at STP.
1 𝑚𝑜𝑙 𝑁2 2 𝑚𝑜𝑙 𝑁𝐻3 17.04 𝑔 𝑁𝐻3
5.00 𝐿 𝑁2 𝑥
𝑥
𝑥
= 7.61 𝑔 𝑁𝐻3
22.4 𝐿 𝑁2 1 𝑚𝑜𝑙 𝑁2
1 𝑚𝑜𝑙 𝑁𝐻3
N2 (g) + 3 H2 (g) → 2 NH3 (g)
• Find the mass of ammonia produced if 5.00 L of nitrogen gas
reacts with excess hydrogen gas at 3.00 atm and 25 oC. (Hint:
you have to do the stoichiometry AND use the Ideal Gas Law
because this is not at STP. The important thing is that you need
to moles of nitrogen!)
• V = 5.00 L P = 3.00 atm
T = 25oC
n=?
• PV=nRT
• (3.00
𝐿 𝑎𝑡𝑚
atm)(5.00L)=n(0.0821
)(298
𝑚𝑜𝑙 𝐾
• n = 0.61 mol N2
K)
• 0.61 𝑚𝑜𝑙 𝑁2 𝑥
2 𝑚𝑜𝑙 𝑁𝐻3
1 𝑚𝑜𝑙 𝑁2
17.04 𝑔𝑁𝐻3
𝑥
1 𝑚𝑜𝑙 𝑁𝐻3
= 21 𝑔 𝑁𝐻3