Answer:
The correct answer is is option B
b. 93.3 g
Explanation:
SEE COMPLETE QUESTION BELOW
Hydrogen chloride gas can be prepared by the following reaction: 2NaCl(s) + H2SO4(aq) → 2HCl(g) + Na2SO4(s)
How many grams of HCl can be prepared from 2.00 mol H2SO4 and 2.56 mol NaCl?
a. 7.30 g
b. 93.3 g
c. 146 g
d. 150 g
e. 196 g
CHECK THE ATTACHMENT FOR STEP BY STEP EXPLANATION
The ratio of moles of reactants to moles of products can be seen from the coefficients in a balanced equation. In our case 4 moles of hydrochloric acid reacts with one mole of oxygen to produce two moles of chlorine and water. So, <span> the ratio of moles of hydrochloric acid to moles of chlorine is 2:1. To determine the number moles, divide the mass by the mass of one mole. </span>
<span>Cl2 = 2 * 35.45 = 70.9 grams </span>
<span>Number of moles = 335 ÷ 70.9 </span>
<span>This is approximately 4.72 moles. The number of moles of hydrochloric acid is twice this number. </span>
<span>Mass of one mole = 1 + 35.46 = 36.45 grams </span>
<span>Total mass = 2 * (335 ÷ 70.9) * 36.45 </span>
<span>This is approximately 344.45 grams.
Correct answer A.</span>
m = given mass of gas = 3.82 g
M = molar mass of gas = ?
T = temperature of laboratory = 302 K
P = air pressure = 1.04 atm = 1.04 x 101325 pa
V = volume of gas = 0.854 L = 0.854 x 10⁻³ m³
using the ideal gas equation
PV = (m/M) RT
inserting the above values
(1.04 x 101325) (0.854 x 10⁻³) = (3.82/M) (8.314) (302)
M = 106.6 g
hence the molar mass of the gas comes out to be 106.6 g
When the concentration is expressed in molality, it is expressed in moles of solute per kilogram of solvent. Since we are given the mass of the solvent, which is water, we can compute for the moles of solute NaNO3.
0.5 m = x mol NaNO3/0.5 kg water
x = 0.25 mol NaNO3
Since the molar mass of NaNO3 is 85 g/mol, the mass is
0.25 mol * 85 g/mol = 21.25 grams NaNO3 needed
