Type your question here 20) Given the 2 generate .25 L following react atm and 1

Question

Type your question here

20) Given the 2 generate .25 L following react atm and 17 0Con, what volume of 6.0 M H2SO4 solution is necessary to ? 2 Al(s) + 3 H2504(aq) - > Al2(SO4)3(aq) + 3 H2(g) 21) What volume of natural gas (CH4) at 1 atm and 22 degree C must burn to emit 500 kj of heat? CH4(g) + 2 02(g) - > 4 CO2(g) + H20(g); AH degree rxn = -802.3 kJ 22) A student mixes 3.07 g of lithium carbonate into 50 mL of a 1.5 M solution of perchloric acid and a gas is formed at 1 atm and 20 degree C. Identify the gas and calculate the work done on the system by the formation of the gas.

Explanation / Answer

3Al + 3H2SO4 --------------> Al2(SO4)3 + 3H2

PV = nRT

Given, P = 0.95 atm

V = 2.25 L

n = ?

R =0.0821 Latm mol^-1 K^-1

T = = 17 degree C = 290 K

(0.95 ) ( 2.25) = n (0.0821)(290)

=> n = 0.0898 mol

So, the moles of H2 gas necessary are 0.0898 mol.

As, the H2SO4 and H2 are in equimolar ratio ( as observed in the chemical reaction),

the moles of H2SO4 necessary are 0.0898 mol in order to produce 0.0898 mol.

So, Moles = Molarity x Volume

=> Volume = Moles / Molarity

=> Volume = 0.0898 mol / 60. M

=> Volume = 0.0196 L

=> Volume = 14.96 mL

------------------------------------------------------------------------------------

21)

From the given chemical reaction, it is evident that, 1 mol of CH4 produces 802.3 kJ of heat.

So, we need to find how many moles of CH4 produces 500 kJ of energy?

So, moles of CH4 = 500 / 802.3

Moles = 0.0623 mol of CH4 is required.

Now, it is given,

P = 1 atm

T = 22 +273 = 295 K

V = ?

n = 0.0623 mol

R = gas constant

PV = nRT (plugging the values)

(1) ( V) = 0.0623 (0.0821)(295)

=> V = 1.51 L

------------------------------------------------------------------------------------------------------

22)

The chemical reaction is

Li2CO3 + 2HClO4 ----------------> 2LiClO4 + H2O + CO2

given, 3.07 g Li2CO3

Moles of Li2CO3 = 3.07 g / 73.8 g/mol = 0.042 mol

Moles of perchloric acid = Molarity x Volume(in L) = 1.5 M x 0.05

Moles = 0.075 mol

Finding limiting reagent.

We need to divide the moles with stoichiometric coefficients

Li2CO3 = 0.042 mol / 1 = 0.042 mol

HClO4 = 0.075 mol / 2 = 0.0375 mol

So, aong them HClO4 has lowest number of moles.

So, it is the limiting reagent.

So, from the chemical reaction, 2 moles of perchloric acid give 1 mol CO2.

So, 0.075 mol of perchloric acid gives 0.0375 mol CO2.

Given,

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.