Hydrogen chloride gas can be prepared by the following reaction: 2NaCl(s) + H2SO4(aq) → 2HCl(g) + Na2SO4(s) How many grams of HC
l can be prepared from 2.00 mol H2SO4 and 2.56 mol NaCl?
1 answer:
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
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Answer:
154.0831 g
Explanation:
The formula for the calculation of moles is shown below:
Given: For
Given mass = 41.0 g
Molar mass of = 24.31 g/mol
Moles of = 41.0 g / 24.31 g/mol = 1.6865 moles
Given: For
Given mass = 175 g
Molar mass of = 162.2 g/mol
Moles of = 175 g / 162.2 g/mol = 1.0789 moles
According to the given reaction:
3 moles of Mg react with 2 moles of
1 mole of Mg react with 2/3 moles of
1.6865 mole of Mg react with (2/3)*1.6865 moles of
Moles of = 1.1243 moles
Available moles of = 1.0789 moles
Limiting reagent is the one which is present in small amount. Thus, is limiting reagent. (1.0789 < 1.1243)
The formation of the product is governed by the limiting reagent. So,
2 moles of gives 3 moles of magnesium chloride
1 mole of gives 3/2 moles of magnesium chloride
1.0789 mole of gives (3/2)*1.0789 moles of magnesium chloride
Moles of magnesium chloride = 1.61835 moles
Molar mass of magnesium chloride = 95.21 g/mol
Mass of magnesium chloride = Moles × Molar mass = 1.61835 × 95.21 g = 154.0831 g
Answer:
324.18 g/mol
Explanation:
Let the molecular mass of the antimalarial drug, Quinine is x g/mol
According to question,
Nitrogen present in the drug is 8.63% of x
So, mass of nitrogen =
Also, according to the question,
2 atoms are present in 1 molecule of the drug.
Mass of nitrogen = 14.01 amu = 14.01 g/mol (grams for 1 mole)
So, mass of nitrogen = 14.01×2 = 28.02
These 2 must be equal so,
solving for x, we get:
<u>x = 324.18 g/mol</u>