$$2H_2 + O_2 → 2H_2O$$
From the equation, we can see that 2 moles of hydrogen react with 1 mole of oxygen to produce 2 moles of water.
First, we need to convert the given masses of oxygen and hydrogen to moles:
$$n_{O_2} = \frac{34 \text{ g}}{32 \text{ g/mol}} = 1.0625 \text{ mol}$$
$$n_{H_2} = \frac{6.0 \text{ g}}{2 \text{ g/mol}} = 3.0 \text{ mol}$$
Comparing the mole ratio of oxygen and hydrogen with the stoichiometric ratio, we can see that hydrogen is in excess. Therefore, we will use oxygen as the limiting reactant to calculate the amount of water produced.
$$n_{H_2O} = 2n_{O_2} = 2 \times 1.0625 \text{ mol} = 2.125 \text{ mol}$$
Now, we can convert the moles of water to liters using the ideal gas law at STP (standard temperature and pressure):
$$PV = nRT$$
At STP, the temperature is 273 K and the pressure is 1 atm. The ideal gas constant is R = 0.08206 L atm/mol K.
$$V_{H_2O} = \frac{n_{H_2O}RT}{P} = \frac{2.125 \text{ mol} \times 0.08206 \text{ L atm/mol K} \times 273 \text{ K}}{1 \text{ atm}}$$
$$V_{H_2O} = 48.6 \text{ L}$$
Therefore, 34 grams of oxygen gas and 6.0 grams of hydrogen at STP can produce 48.6 liters of water.
