Consider the following reaction:
PCl3 + 3H2O -----> P(OH) + 3HCl
if 4.00L of PCl3 react with an excess of H2O at STP, how many grams of HCl would be formed?
Here is an example.
http://www.jiskha.com/science/chemistry/stoichiometry.html
To find out how many grams of HCl would be formed, we need to first calculate the amount of PCl3 that reacts using the given volume and the conditions of STP (Standard Temperature and Pressure).
STP conditions are defined as a temperature of 273.15 K (0°C) and a pressure of 1 atmosphere (atm).
1 mole of any gas at STP occupies 22.4 liters. Since we have 4.00 liters of PCl3, we can convert it to moles by dividing it by the molar volume at STP:
4.00 L PCl3 * (1 mol PCl3 / 22.4 L) = 0.179 moles PCl3
Now, looking at the balanced chemical equation:
PCl3 + 3H2O → P(OH) + 3HCl
We can see that the stoichiometric ratio between PCl3 and HCl is 1:3. This means that for every 1 mole of PCl3, we will produce 3 moles of HCl.
Since we have 0.179 moles of PCl3, we can calculate the amount of moles of HCl produced:
0.179 moles PCl3 * (3 moles HCl / 1 mole PCl3) = 0.537 moles HCl
Finally, we need to convert the moles of HCl to grams. The molar mass of HCl is approximately 36.46 g/mol.
0.537 moles HCl * 36.46 g/mol = 19.8 grams of HCl
Therefore, approximately 19.8 grams of HCl would be formed when 4.00 liters of PCl3 react with an excess of H2O at STP.