365g of hydrochloric acid reacts with 230g of sodium metal in a controlled environment. Calculate the number of moles and volume of hydrogen gas liberated at STP.
Answers
Answer:
This is a “single state” problem, so we can solve it using the ideal gas law, PV = nRT. In order to find the volume of hydrogen gas (V), we need to know the number of moles of hydrogen that will be produced by the reaction. Our stoichiometry is simply one mole of hydrogen per mole of zinc, so we need to know the number of moles of zinc that are present in 5.98 grams of zinc metal. The temperature is given in centigrade, so we need to convert into Kelvin, and we also need to convert mm Hg into atm.
Conversions:
25.0C+273=298K(9.6.1)
(742mmHg)×(1atm760mmHg)=0.976atm(9.6.2)
(5.98gZn)×(1.00mol65.39gZn)=0.0915mol(9.6.3)
Substituting:
PV=nRT(9.6.4)
(0.976atm)×V=(0.0915mol)(0.0821Latmmol−1K−1)(298K)(9.6.5)
V=(0.0915mol)(0.0821Latmmol−1K−1)(298K)(0.976atm)=2.29L(9.6.6)
The number of moles of hydrogen gas is 5 moles and the volume of hydrogen gas at STP is 112 L.
Given,
Amount of hydrochloric acid = 365gram.
Amount of sodium metal = 230 gram.
To Find,
The number of moles and volume of hydrogen gas extricated at STP.
Solution,
We can solve this type of provided question in the following manner.
First of all, compose a balanced equation:
2HCl + 2Na → 2NaCl + H2
- So by this equation, we can say that 2moles of HCl react with 2moles of sodium to give 2moles of NaCl and 1 mole of hydrogen.
- Takes anyone either HCl or sodium and applies the unitary technique if 2 moles of HCl provide 1 mole of hydrogen then, 1 mole of HCl provides 1/2 moles of hydrogen now HCL is given 365g so evaluate its mole.
Now HCL is given 365g so evaluate its mole, moles of HCL = 365/36.5 ⇒ 10moles of HCl
- So 10 moles of HCl gives 1×10/2 moles of hydrogen ⇒ 5 moles of hydrogen.
To calculate the volume
v = 5×22.4 L v = 112 L
Hence, the number of moles of hydrogen gas is 5 moles and the volume of hydrogen gas at STP is 112 L.
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