How many grams of NH3, are required to produce 5.23 g HF?

Here is the set up for a similar problem I did just a few minutes ago. Just follow the template.

http://www.jiskha.com/display.cgi?id=1270062101

you useless DrBob

To determine the grams of NH3 required to produce a given amount of HF, you need to consider the balanced chemical equation of the reaction between NH3 and HF.

The balanced equation for the reaction between NH3 and HF is:

NH3 + HF -> NH4F

From the equation, we can see that the stoichiometric ratio between NH3 and HF is 1:1. This means that for every 1 mole of NH3, 1 mole of HF is produced.

To calculate the amount of NH3 required to produce a given amount of HF, you can follow these steps:

1. Calculate the moles of HF:
Moles of HF = mass of HF / molar mass of HF

The molar mass of HF can be calculated by adding the atomic mass of hydrogen (H) and fluorine (F). The atomic masses of hydrogen and fluorine are approximately 1 g/mol and 19 g/mol, respectively.

2. Use the stoichiometric ratio to determine the moles of NH3:
Moles of NH3 = Moles of HF

3. Finally, calculate the mass of NH3:
Mass of NH3 = Moles of NH3 * molar mass of NH3

The molar mass of NH3 can be calculated by adding the atomic mass of nitrogen (N) and hydrogen (H). The atomic masses of nitrogen and hydrogen are approximately 14 g/mol and 1 g/mol, respectively.

Following these steps, you can calculate the grams of NH3 required to produce 5.23 g HF.