Question

please explain and help calculate the orange cirlced ones

L in volume at STP 1. One mole of an IDEAL GAS occupies a) 12.2 6) 22.4 c) 23.6 d) 40 (2What are the units for the molar volume of a gas? -a) Moles/liter -b) Liters/mole Grams/mole d) Moles/grams What is the gas produce in this experiment CaCO3+ HCI? a) O2 b) CO2 c) NO2 d) H2 e) Natural gas on: 2:23 los molhoz 4 When CaCO3 react with HCI, calculate the mass of CaCO3 that would produce 50 mL of CO2 at STP. a) 0.18 g PV=nRT -2127 3 p=latm. -b) 0.22 g 2.23 100.79 27 RT ni moja tu 273:15 K C) 0.35 g moloz imel d) 0.50 g na!150) 30 V: 50 m Caloa 0.08206 (273.15 22.41 R=0.08206 (5) A 23.0-ml volume of Oz, collected over water at 752 torr and 22°C, is produced from the thermal decomposition of KCIO3. The vapor pressure of water at 22°C is 19.8 torr. 2 KCIO3(s) — 2 KCl(s) + 3 O2(g) How many moles of O2 are collected? PV=nRT a) 9.15 x 10-mol b) 8.75 x 10-mol c) 4.58 x 10"mol d) 3.15 x 10mol

Answer 1

2 ) Molar volume of gas is volume of gas per mole. Hence, its unit is L / mole.

ANSWER : a) Liter / mole

3) Consider reaction, CaCO ₃ (aq) + 2 HCl (aq)  $\rightarrow$ CaCl ₂ (aq) + H₂O (l) + CO ₂ (g)

ANSWER : b) CO ₂ (g)

4) Consider reaction, CaCO ₃ (aq) + 2 HCl (aq)  $\rightarrow$ CaCl ₂ (aq) + H₂O (l) + CO ₂ (g)

From reaction, 1 mol CaCO ₃$\equiv$ 2 mol HCl $\equiv$ 1 mol CaCl ₂ $\equiv$ 1 mol CO ₂

At STP, volume of 1 mole of gas = 22.4 L .

Hence, we can write 22.4 L CO ₂ gas $\equiv$ 1 mol CO ₂ gas

$\therefore$ 0.050 L CO ₂ gas $\equiv$ 1 $\times$ 0.050 / 22.4 mol CO ₂ gas

0.050 L CO ₂ gas $\equiv$ 0.00223 mol CO ₂ gas

From reaction, 1 mol CaCO ₃$\equiv$ 1 mol CO ₂

Therefore, no. of moles of CaCO ₃ = no. of moles of CO ₂ = 0.00223 mol

We know that , no. of moles = Mass / Molar mass

Molar mass of CaCO ₃ = 40.08 + 12.01 + ( 3 $\times$ 16.00 ) = 100.09 g / mol

$\therefore$ Mass of CaCO ₃ = no. of moles of CaCO ₃  $\times$ Molar mass

$\therefore$ Mass of CaCO ₃ = 0.00223 mol $\times$ 100.09 g / mol = 0.223 g

ANSWER : b) 0.22 g

5)

According to Dalton's law, When gas collected over water then the total pressure of a gas is the sum of the partial pressure of a gas and partial pressure of water.

i e P total = P oxygen + P water vapor.

From given table, P water vapor at 30.0 ⁰ C = 31.8 torr .

P oxygen = P total - P water vapor

P oxygen = 752 torr - 19.8 torr

P oxygen= 732.2 torr

Now, calculate no. of moles of oxygen gas.

We have relation, P V = n R T , where, P is a pressure of a gas, V is a volume of a gas, n is no. of moles of gas, R is a gas constant and T is temperature of gas.

$\therefore$ n = P V / R T

We have , P = 732.2 torr = 0.963 atm , V = 23.0 ml = 0.023 L , T = 22 + 273 = 295 K

$\therefore$ no. of moles of oxygen gas = 0.963 atm $\times$ 0.023 L / ( 0.082057 L atm / mol.K $\times$ 295 K )

$\therefore$ no. of moles of oxygen gas = 9.15 $\times$ 10 ^-04 mol

ANSWER : a) 9.15 $\times$ 10 ^-04 mol