When solid calcium is placed in water, it undergoes a vigorous reaction that produces solid calcium oxide, CaO(s) and hydrogen gas. determine the volume of dry hydrogen gas that is produced when 5.00 g of calcium reacts and the gas is collected at 42 degrees celsius and 770 mmHg.

Ca (s) + H2O (l) = CaO (s) + H2 (g)

I don't agree with the equation because you can't isolate CaO. It immediately reacts with H2O to form CaOH)2 as I've indicated in the equation below. However, none of that changes the stoichiometry of the equation and 1 mol Ca still produces 1 mol H2 gas.

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

mols Ca = grams Ca/atomic mass Ca = ?
1 mol Ca produces 1 mol H2 gas.
That 1 mol H2 gas will be dry H2 gas. Use PV = nRT and solve for V in liters at the conditions listed.

To determine the volume of dry hydrogen gas produced when 5.00 g of calcium reacts, we can use the ideal gas law equation: PV = nRT.

First, we need to convert the mass of calcium (5.00 g) to moles. We can do this by dividing the mass by the molar mass of calcium.

The molar mass of calcium (Ca) is 40.08 g/mol, so:

5.00 g / 40.08 g/mol = 0.1246 mol

Now, let's consider the balanced equation: Ca(s) + H2O(l) → CaO(s) + H2(g)

From the equation, we can see that 1 mole of calcium will produce 1 mole of hydrogen gas (H2). Therefore, the number of moles of hydrogen gas produced is also 0.1246 mol.

Next, we need to convert the conditions from Celsius to Kelvin (K) to use in the ideal gas law equation. The conversion is done by adding 273.15 to the Celsius temperature:

42 °C + 273.15 = 315.15 K

We also need to convert the pressure from mmHg to atm, dividing by 760 (since 760 mmHg is equal to 1 atm):

770 mmHg / 760 mmHg/atm = 1.01 atm

Now, we can plug the values into the ideal gas law equation:

PV = nRT

V = (nRT) / P

V = (0.1246 mol) x (0.0821 L·atm/mol·K) x (315.15 K) / (1.01 atm)

Calculating, we find that:

V ≈ 3.48 L

Therefore, the volume of dry hydrogen gas produced when 5.00 g of calcium reacts is approximately 3.48 liters.