Answer:
Explanation:
We must do the conversions
mass of C₆H₁₂O₆ ⟶ moles of C₆H₁₂O₆ ⟶ moles of CO₂ ⟶ volume of CO₂
We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.
Mᵣ: 180.16
C₆H₁₂O₆ + 6O₂ ⟶ 6CO₂ + 6H₂O
m/g: 24.5
(a) Moles of C₆H₁₂O₆
(b) Moles of CO₂
(c) Volume of CO₂
We can use the Ideal Gas Law.
pV = nRT
Data:
p = 0.960 atm
n = 0.8159 mol
T = 37 °C
(i) Convert the temperature to kelvins
T = (37 + 273.15) K= 310.15 K
(ii) Calculate the volume
Answer:
113.8g
Explanation:
Statement of problem: mass of 1.946mole of NaCl
Given parameters:
Number of moles of NaCl = 1.946mole
Unknown: mass of NaCl
Solution
To find the mass of NaCl, we apply the concept of moles which expresses the relationship between number of moles and mass according to the equation below:
Number of moles = 
To find the molar mass of NaCl:
the atomic mass of Na = 23g
atomic mass of Cl = 35.5g
Molar mass of NaCl = (23 + 35.5) = 58.5gmol⁻¹
Mass of NaCl = Number of moles x molar mass of NaCl
Mass of NaCl = 1.946 x 58.5 = 113.8g
The mass of a NaCl solution that is required to prepare 0.40 L of a 0.75 M solution is 17.55g. Details about mass can be found below.
<h3>How to calculate mass?</h3>
The mass of a substance can be calculated by multiplying the number of moles by its molar mass.
However, the number of moles of a solution must be initially calculated by using the following formula:
molarity = no of moles ÷ volume
no of moles = 0.75 × 0.40
no of moles = 0.3 moles
mass of NaCl = 0.3 × 58.5 = 17.55g
Therefore, the mass of a NaCl solution that is required to prepare 0.40 L of a 0.75 M solution is 17.55g.
Learn more about mass at: brainly.com/question/19694949
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Answer:
Theoretical yield of the reaction = 34 g
Excess reactant is hydrogen
Limiting reactant is nitrogen
Explanation:
Given there is 100 g of nitrogen and 100 g of hydrogen
Number of moles of nitrogen = 100 ÷ 28 = 3·57
Number of moles of hydrogen = 100 ÷ 2 = 50
Reaction between nitrogen and hydrogen yields ammonia according to the following chemical equation
N2 + 3H2 → 2NH3
From the above chemical equation for every mole of nitrogen that reacts, 3 moles of hydrogen will be required and 2 moles of ammonia will be formed
Now we have 3·57 moles of nitrogen and therefore we require 3 × 3·57 moles of hydrogen
⇒ We require 10·71 moles of hydrogen
But we have 50 moles of hydrogen
∴ Limiting reactant is nitrogen and excess reactant is hydrogen
From the balanced chemical equation the yield will be 2 × 3·57 moles of ammonia
Molecular weight of ammonia = 17 g
∴ Theoretical yield of the reaction = 2 × 3·57 × 17 = 121·38 g
