magnesium react with hydrochloric acid,what is the volume of hydrogen produced at r.t.p when 0.12g of magnesium ribbon react?
balance the equation
Mg+2HCl>>>MgCl2 + H2
so one gets the same number of moles of H2 as moles one used of Mg.
MolesMg=.12/atomicmassMg
volumeH2=molesH2*RTPmolarvolume
Volume of hydrogen is 24dm3
To calculate the volume of hydrogen produced when 0.12g of magnesium ribbon reacts with hydrochloric acid at RTP (room temperature and pressure), you will need to use the stoichiometry of the balanced chemical equation. The balanced chemical equation for the reaction between magnesium and hydrochloric acid is:
Mg + 2HCl -> MgCl2 + H2
From the equation, we can see that 1 mole of magnesium produces 1 mole of hydrogen gas.
1 mole of magnesium is equal to its molar mass, which is 24.31 grams.
First, we need to determine the number of moles of magnesium in 0.12g:
0.12g / 24.31 g/mol = 0.00493 moles
Since the stoichiometry of the reaction is 1:1 (1 mole of magnesium produces 1 mole of hydrogen gas), the number of moles of hydrogen gas produced will also be 0.00493 moles.
Next, we can use the ideal gas law to calculate the volume of hydrogen gas produced at RTP (room temperature and pressure). The conditions for RTP are 1 atmosphere of pressure and 0 degrees Celsius (or 273 Kelvin).
The ideal gas law equation is: PV = nRT
where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin.
At RTP, the pressure is 1 atmosphere and the temperature is 273 Kelvin.
Plugging in the values:
(1 atm) * V = (0.00493 moles) * (0.0821 L*atm/mol*K) * (273 K)
V = (0.00493 mol * 0.0821 L*atm/mol*K * 273 K) / (1 atm)
V = 0.112 L
Therefore, the volume of hydrogen gas produced at RTP when 0.12g of magnesium ribbon reacts is approximately 0.112 liters.
To calculate the volume of hydrogen gas produced when magnesium reacts with hydrochloric acid, you need to apply the principles of stoichiometry and the ideal gas law.
First, let's write the balanced chemical equation for the reaction between magnesium and hydrochloric acid:
Mg + 2HCl → MgCl2 + H2
According to the stoichiometry of the reaction, one mole of magnesium (24.31 g) reacts with 2 moles of hydrochloric acid to produce 1 mole of hydrogen gas.
To find the moles of magnesium present in 0.12 g of magnesium ribbon, you can use the formula:
moles = mass / molar mass
The molar mass of magnesium is 24.31 g/mol, so:
moles of magnesium = 0.12 g / 24.31 g/mol
Now, since the stoichiometry tells us that the ratio between magnesium and hydrogen gas is 1:1, we can conclude that the moles of hydrogen gas produced will be the same as the moles of magnesium.
Next, we need to use the ideal gas law to determine the volume of hydrogen gas at RTP (room temperature and pressure). RTP conditions are defined as 1 atm of pressure and 273.15 K (0°C).
The ideal gas law is expressed as:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in L)
n = moles of gas
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature (in Kelvin)
Since we want to find the volume (V) of hydrogen gas, rearrange the equation to solve for V:
V = (nRT) / P
Substituting the values:
n = moles of hydrogen = moles of magnesium
R = 0.0821 L·atm/mol·K
T = 273.15 K
P = 1 atm
Now you have all the necessary information to calculate the volume of hydrogen gas produced at RTP. Just plug in the values and calculate:
V = (moles of magnesium) x (0.0821 L·atm/mol·K) x (273.15 K) / 1 atm
Remember to provide the number of significant figures appropriate for the given data.