Hydrogen cyanide is produced industrially from the reaction of gaseous ammonia, oxygen, and methane.

2 NH3(g) + 3 O2(g) + 2 CH4(g) �¨ 2 HCN(g) + 6 H2O(g)
If 5.27 103 kg each of NH3, O2, and CH4 are reacted, what mass of H2O will be produced, assuming 100% yield?

that would be 5.27 times 10 to the third kg by the way.

Hydrogen cyanide is produced industrially from the reaction of gaseous ammonia, oxygen, and methane.

2 NH3(g) + 3 O2(g) + 2 CH4(g) �¨ 2 HCN(g) + 6 H2O(g)
If 5.27x10^3 kg each of NH3, O2, and CH4 are reacted, what mass of HCN and of H2O will be produced, assuming 100% yield?

confused

To find the mass of water (H2O) produced, we need to determine the molar ratio of water to ammonia (NH3) in the balanced equation.

From the balanced equation, we can see that for every 2 moles of NH3, 6 moles of H2O are produced. Therefore, the molar ratio of H2O to NH3 is 6:2, which simplifies to 3:1.

To solve this problem, we will follow these steps:

Step 1: Find the moles of NH3

The molar mass of NH3 (ammonia) is:
(1 x 14.01 g/mol) + (3 x 1.01 g/mol) = 17.03 g/mol

Given the mass of NH3 as 5.27 x 10^3 kg, we can convert it to grams:
5.27 x 10^3 kg = 5.27 x 10^6 g

To find the moles of NH3, we divide the mass by the molar mass:
moles of NH3 = 5.27 x 10^6 g / 17.03 g/mol

Step 2: Calculate the moles of H2O produced

Since the molar ratio of H2O to NH3 is 3:1, the moles of H2O produced is three times the moles of NH3:
moles of H2O = 3 x moles of NH3

Step 3: Convert the moles of H2O to grams

The molar mass of H2O (water) is:
(2 x 1.01 g/mol) + (16.00 g/mol) = 18.02 g/mol

To find the mass of H2O, we multiply the moles of H2O by the molar mass:
mass of H2O = moles of H2O x molar mass

Let's calculate the mass of H2O:

moles of NH3 = (5.27 x 10^6 g) / (17.03 g/mol)
moles of H2O = 3 x moles of NH3
mass of H2O = moles of H2O x molar mass of H2O

By plugging in the values, we can calculate the mass of H2O.