the volume of nh3 at stp needed to pass into 30 ml of in h2so4 solution to bring down its strength to m/10
what is m/10?
To determine the volume of NH3 gas at STP (Standard Temperature and Pressure) required to bring down the strength of an H2SO4 solution to m/10, we need to consider the balanced chemical equation for the reaction between NH3 and H2SO4.
The balanced chemical equation is:
2NH3 + H2SO4 → (NH4)2SO4
From the equation, we can see that two moles of NH3 react with one mole of H2SO4 to form one mole of (NH4)2SO4.
Given that the solution needs to be diluted to m/10, which means one-tenth of the original concentration, we can set up the following equation based on the relationship between moles, concentration, and volume:
(Moles of H2SO4 initially) / (Initial Volume of H2SO4) = (Moles of H2SO4 after dilution) / (Final Volume of Solution)
Now, let's calculate the moles of H2SO4 initially in the solution.
Molarity (M) = Moles / Volume (in liters)
Converting 30 ml to liters: 30 ml = 30/1000 = 0.03 L
Since the concentration is m/10, the moles of H2SO4 initially can be found using the molarity formula:
(M/10) = Moles / 0.03 L
Rearranging the equation, we have:
Moles of H2SO4 initially = (M/10) * 0.03
Since the stoichiometry of the reaction tells us that 2 moles of NH3 are needed to react with 1 mole of H2SO4, we can calculate the moles of NH3 required using the ratio:
Moles of NH3 = 2 * (Moles of H2SO4 initially)
Now, to find the volume of NH3 gas at STP, we'll use the ideal gas law:
PV = nRT
Since we have the moles of NH3, we can rearrange the formula to solve for volume (V):
V = (nRT) / P
At STP, the pressure (P) is 1 atmosphere, the temperature (T) is 273.15 Kelvin, and the ideal gas constant (R) is 0.0821 L•atm/(mol•K).
Plugging in the values, we get:
V = (moles of NH3 * 0.0821 * 273.15) / 1
Finally, you can substitute the calculated value for "moles of NH3" into the equation to find the volume of NH3 gas at STP needed to perform the desired dilution.