Question

Каков объём водорода (в л, н.у.) образуется при взаимодействии 12 г кальция с избытком воды? Ответ запишите с точностью до сотых, без указания единиц измерения.

Answer 1

Ca(0,3)+2H2O->Ca(OH)2+H2(0,3)

n(Ca)=12:40=0,3моль

V(H2)=0,3×22,4=6,72литра

Answer 2

Calculation of Hydrogen Volume

To calculate the volume of hydrogen gas formed when 12 grams of calcium reacts with an excess of water, we need to consider the balanced chemical equation for the reaction:

Ca + 2H2O -> Ca(OH)2 + H2

From the equation, we can see that 1 mole of calcium reacts with 2 moles of water to produce 1 mole of hydrogen gas. To determine the number of moles of calcium, we can use its molar mass, which is 40.08 g/mol.

The number of moles of calcium can be calculated as follows:

moles of calcium = mass of calcium / molar mass of calcium
                = 12 g / 40.08 g/mol
                = 0.2997 mol (rounded to four decimal places)

Since the reaction is 1:2 between calcium and water, the number of moles of water is twice the number of moles of calcium:

moles of water = 2 * moles of calcium
              = 2 * 0.2997 mol
              = 0.5994 mol (rounded to four decimal places)

Therefore, the number of moles of hydrogen gas produced is also equal to the number of moles of calcium:

moles of hydrogen gas = moles of calcium
                     = 0.2997 mol (rounded to four decimal places)

To convert the number of moles of hydrogen gas to volume, we can use the ideal gas law, which states that:

PV = nRT

Where: P = pressure (in atm) V = volume (in liters) n = number of moles R = ideal gas constant (0.0821 L·atm/(mol·K)) T = temperature (in Kelvin)

Assuming standard temperature and pressure (STP), which is 0 degrees Celsius (273.15 Kelvin) and 1 atm, respectively, we can calculate the volume of hydrogen gas:

V = nRT / P
  = 0.2997 mol * 0.0821 L·atm/(mol·K) * 273.15 K / 1 atm
  = 6.949 L (rounded to three decimal places)

Therefore, the volume of hydrogen gas formed when 12 grams of calcium reacts with an excess of water is approximately 6.949 liters.