The balanced equation for the reaction between Iodine and Hydrogen is:
H2 + I2 --> 2HI
The ratio of I2 to HI is 1:2
Therefore 1 mole of Iodine can form a maximum of 2 moles of Hydrogen Iodide
Chat with our AI personalities
When 1 mole of iodine reacts with excess hydrogen according to the balanced equation: I2 + H2 → 2HI, 1 mole of hydrogen iodide (HI) is produced.
Since molecules of potassium contain only single potassium atoms, molecules of iodine contain two atoms, and moles of potassium iodide contain one atom of each element, 2.5 moles of iodine are needed to react completely with 5 moles of potassium.
If 0.36 moles of zinc reacts with an equal amount of H2SO4, then the stoichiometry of the reaction tells us that for every 1 mole of zinc, 1 mole of hydrogen gas is produced. Therefore, 0.36 moles of hydrogen gas will be produced.
To find the number of grams in 0.02 moles of beryllium iodide (BeI2), you would first calculate the molar mass of BeI2, which is 262.83 g/mol. Then, you would multiply the molar mass by the number of moles: 0.02 moles * 262.83 g/mol = 5.26 grams of beryllium iodide.
The mole ratio of hydrogen to ammonia in the reaction is 3:2. This means that for every 3 moles of hydrogen that react, 2 moles of ammonia are produced.
To find the volume of hydrogen gas produced, we first need to convert the mass of baking soda (645g) to moles. Then, using the balanced chemical equation for the reaction, we can determine the moles of hydrogen gas produced. Finally, using the ideal gas law at STP, we can convert the moles of hydrogen gas to liters.