The Haber process can be used to produce ammonia (NH3) from hydrogen gas (H2) and nitrogen gas (N2). The balanced equation for this process is shown below.3H2 + N2 mc025-1.jpg 2NH3
The molar mass of NH3 is 17.03 g/mol. The molar mass of H2 is 2.0158 g/mol. In a particular reaction, 0.575 g of NH3 forms. What is the mass, in grams, of H2 that must have reacted, to the correct number of significant figures?
0.1
0.102
0.10209
0.1021
Answers
Answer : The mass of in grams is 0.102g.
Solution : Given,
Molar mass of = 17.03 g/mole
Molar mass of = 2.0158 g/mole
Given Mass of = 0.575 g
First we have to calculate the moles of .
=
= 0.0337 moles
The given balanced equation is,
From the above reaction, we conclude that
2 moles of produced from 3 moles of
then the 0.0337 moles of produces to give
moles of
The moles of = 0.0505 moles
The mass of = Moles of
× Molar mass of
= 0.0505 moles × 2.0158 g/mole = 0.10179 g
The mass of in the correct number of significant figures is 0.102 g.
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Answers
Answer: chlorine gas
Explanation:
mass = 0.508 g volume = 0.175 L
From ideal gas equation.
PV = nRT
(1.0 atm) × (0.175 L) = n × [0.0821 atm L / (mol K)] × [(273.2 + 25.0) K]
n = 1.0 × 0.175 / (0.0821 × 298.2) mol
n = 0.00715 mol
Molarmass= mass/mole
Molar mass of the gas = (0.508 g) / (0.00715 mol) = 71.0 g/mol
The molar mass of Cl₂ (chorine) = 35.5 × 2 g/mol = 71.0 g/mol
Hence, the gas is chlorine.
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Final answer:
When 150 ml of 0.500 M silver nitrate are added to 100 mL of 0.400 M potassium chromate, a silver chromate precipitate forms. Considering the stoichiometry of the reaction and the quantities of reactants, 24.88 grams of silver chromate will precipitate.
Explanation:
The subject of this question is based on precipitation reactions in Chemistry. Precipitation reactions occur when two solutions combine to form an insoluble solid known as a precipitate. The moles of silver nitrate present in a 150 mL of 0.500 M solution can be calculated using the formula Molarity = Moles ÷ Volume (in Litres).
Thus, Moles of AgNO3 = 0.500 M * 0.15 L = 0.075 mol AgNO3. According to the reaction equation 2AgNO3 + K2CrO4 → 2AgCrO4(precipitate) + 2KNO3, for every mole of K2CrO4, we have two moles of AgNO3. Thus, based on stoichiometry and the given quantities of the reactants, the limiting reactant will be AgNO3, and it will totally react and form the silver chromate precipitate. The moles of Ag2CrO4 formed would therefore also be 0.075 mol. To convert this into grams, we use the molar mass of Ag2CrO4, which is approximately 331.73 g/mol. Hence, grams of Ag2CrO4 = 0.075 mol Ag2CrO4 * 331.73 g/mol = 24.88 g Ag2CrO4.
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The controlled fusion reaction is a nuclear reaction in which two or more atomic nuclei come very close and then collide at a very high speed and join to form a nucleus. The mass of the new nucleus is slightly less than the sum of its components b an amount known as binding energy.