<em>What volume do 5 moles of a gas occupy at 28 ° C and 3 atm of pressure?</em>
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<h3>Further explanation</h3>In general, the gas equation can be written
where
P = pressure, atm
V = volume, liter
n = number of moles
R = gas constant = 0.08206 L.atm / mol K
T = temperature, Kelvin
n= 5 moles
T=28 +273=301 K
P=3 atm
The volume of the gas :
Answer:
0.4 g of hydrogen gas would be produced.
1. 0.48 mole of HCl is needed to react completely with 15.5 g of zinc.
2. HCl is the limiting reactant.
3. 2.61 g of zinc is in excess
Explanation:
From the balanced equation of reaction:
1 mole of Zn requires 2 moles of HCl to produce 1 mole of hydrogen gas.
15.5 g of zinc = 15.5/65.3 = 0.24 moles of zinc.
0.24 moles Zn is supposed to require 0.24 x 2 moles HCl which is equivalent to 0.48 moles HCl.
But only 0.400 mole of HCl is present. <u>Hence, HCl is the limiting reagent</u>. <u>For complete reaction with 15.5 g of Zinc, 0.48 mole HCl would be needed.</u>
0.400 mole of HCl will require 0.2 mole of Zn for complete reaction. <u>This thus means that 0.24 - 02 = 0.04 mole of Zn is in excess.</u>
0.04 mole Zn = 0.04 x 65.3 = 2.61 g excess Zn.
Now, since HCl is the limiting reagent;
2 moles of HCl is required to produce 1 mole of H2 according to the equation.
0.400 mole HCl will therefore yield 0.400 x 1/2 = 0.2 mole H2
0.2 mole H2 = 2 x 0.2 = 0.4 g H2
<em>Hence, </em><em>0.4 g</em><em> of hydrogen gas would be produced.</em>
Answer:
0.252 mol
Explanation:
<em>Given the following reaction: </em>
<em>Cu + 2 AgNO₃ → 2 Ag + Cu(NO₃)₂</em>
<em>How many moles of Ag will be produced from 16.0 g Cu, assuming AgNO3 is available in excess.</em>
First, we write the balanced equation.
Cu + 2 AgNO₃ → 2 Ag + Cu(NO₃)₂
We can establish the following relations.
The moles of Ag produced from 16.0 g of Cu are: