Question

Need help answeringg questions; 14 and 15. Thanks

14. Consider the reaction MnO2 (s) + 2Cl(aq) + 4H+(aq) - Mn+2 (aq) + CI, (g) + 2 H2O( How many grams of manganese dioxide are required to produce 1.2 L of chlorine gas at 1.0 atm and 200 deg C? 15. A 200 mL sample of oxygen is collected over water at 27 deg C and 748 mmHg. The vapor pressure of water at 27 deg C is 28 mmHg. What mass of oxygen is in the sample?

Answer 1

#14.

As the balanced chemical equation shows that 1mol of Cl2(g) is produced by 1mol of MnO2(s).

Lets calculate the moles of Cl2(g) produced from given data by using ideal gas equation;

PV = nRT

P = pressure of gas = 1atm

V = liter volume of gas = 1.2L

n = moles of gas , to be calculated here.

R = gas constant = 0.0821Latm/mol.K

T = Kelvin temperature, 200°C = 200+273.15 = 473.15K

Using all values to calculate (n)

n = (PV)÷(RT)

n = (1atm × 1.2L) ÷ (0.0821Latm/mol.K × 473.15K)

n = 0.0309 mol of Cl2 (g)

Then moles of MnO2 must be required = 0.0309mol

Mass of MnO2 required = moles × molar mass = 0.0309mol × 86.94g/mol = 2.70g

Thus mass of MnO2 required = 2.70g

#15.

Pressure due to oxygen only = total pressure - pressure of water = 748mmHg - 28mmHg = 720mmHg × (1atm/760mmHg) = 0.9474atm

Volume of oxygen = 200mL × (0.001L/1mL) = 0.200L

Temperature = 27°C = 27 +273.15 = 300.15K

Gas constant (R) = 0.0821L.atm/mol.K

Use all values to calculate moles of oxygen gas as ;

n = (0.9474atm × 0.200L) ÷ (0.0821L.atm/mol.K × 300.15K)

n = 0.0077mol of oxygen gas

Mass of oxygen (O2) gas = moles × molar mass = 0.0077mol × 32g/mol = 0.25g of O2 gas .