Match the grams of each element or compound to the corresponding number of moles of that element or compound.
1. 20.0g h 0.011 moles
2. 0.50 g na 20.0 moles
3. 2.0 x 102 mg h2o 0.022 moles

Answer is: 20.0g H - 20.0 moles; 0.50 g Na - 0.022 moles; 2.0 x 102 mg H2O - 0.011 moles.

1) m(H) = 20.0 g; mass of hydrogen.

n(H) = m(H) ÷ M(H).

n(H) = 20 g ÷ 1 g/mol.

n(H) = 20.0 mol; amount of hydrogen.

2) m(Na) = 0.50 g; mass of sodium.

n(Na) = m(Na) ÷ M(Na).

n(Na) = 0.5 g ÷ 22.98 g/mol.

n(Na) = 0.022 mol; amount of sodium.

3) m(H₂O) = 200 mg ÷ 1000 mg/g.

m(H₂O) = 0.20 g; mass of water.

n(H₂O) = m(H₂O) ÷ M(H₂O).

n(H₂O) = 0.2 g ÷ 18 g/mol.

n(H₂O) = 0.011 mol; amount of water.

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :