Answer:
- <em>You could expect 3.48 grams of C₂H₄N₂</em>
Explanation:
You must start by stating the chemical equation for the reaction of ammonia, carbon dioxide, and methane to produce aminoaceto nitrile.
1. Word equation:
Ammonia + Carbon dioxide + Methane → Aminoacetonitrile + Water
2. Balanced chemical equation:
3. Convert the mass of each reactant into number of moles:
<u>Formula:</u>
- Number of moles = mass in grams/molar mass
<u>2.11g NH₃</u>
- Number of moles = 2.11g / 17.03g/mol = 0.124 mol NH₃
<u>14.9g CO₂</u>
- Number of moles = 14.9g/44.01g/mol = 0.339 mol CO₂
<u>1.75g CH₄</u>
- Number of moles = 1.75g/16.04g/mol = 0.109 mol CH₄
4. Theoretical mol ratio
From the balanced chemical equation, using the coefficientes:
5. Limiting reagent
The available amounts of the reactants are:
Fom the theoretical mole ration, to react with 0.124 mol of NH₃ you would need:
- 0.124molNH₃ × (5molCO₂/8molNH₃) = 0.0775 mol CO₂
Since there are 0.339 moles available, this is in excess.
- 0.124molNH₃ × (3molCH₄/8molNH₃) = 0.0465mol CO₂
Since there are 0.109 moles available, this is in excess too.
Hence, the limiting reagent is NH₃.
6. Yield
Use the theoretical ratio:
- 0.124molNH₃ × (4molC₂H₄N₂ / 8molNH₃) = 0.0620 mol C₂H₄N₂
Convert to grams:
- Mass = number of moles × molar mass
- 0..0620 mol × 56.068g/mol = 3.48 g of C₂H₄N₂ ← answer
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Answer:
30 mL VOLUME OF 3.0 M HCl SHOULD BE USED BY THE STUDENT TO MAKE A 1.80 M IN 50 mL OF HCl.
Explanation:
M1 = 3.00 M
M2 = 1.80 M
V2 = 50 .0 mL = 50 /1000 L = 0.05 L
V1 = unknown
In solving this question, we know that number of moles of a solution is equal to the molar concentration multiplied by the volume. To compare two samples, we equate both number of moles and substitute for the required component.
So we use the equation:
M1 V1 = M2 V2
V1 = M2 V2 / M1
V2 = 1.80 * 0.05 / 3.0
V2 = 0.09 /3.0
V2 = 0.03 L or 30 mL
To prepare the sample of 1.80 M HCl in 50.0 mL from a 3.0 M HCl, 30 mL volume should be used.
The Relative Formula Mass of NaH2PO4 is 120 g/mol
Therefore, the number of moles = 6.6/120
= 0.055 moles of NaH2PO4 which is also equal to the number of moles of H2PO4.
[H2PO4-] = Number of moles oof H2PO4-/Volume of the solution in L
= 0.055/ ( 355 ×10^-3)
= 0.155 M
Na2HPO4 undergoes complete dissociation as follows;
Na2HPO4 (aq)= 2Na+ (aq) + HPO4^2- (aq)
1 mole of Na2HPO4 = 142 g/mol
Therefore; number of moles = 8.0/142
= 0.0563 moles
[HPO4 ^-2] is given by no of moles HPO4^2- /volume of the solution in L
= 0.0563/(355×10^-3)
= 0.1586 M
Both H2PO4^2- and HPO4^2- are weak acids the undergoes partial dissociation
Ka of H2PO4- = 6.20 × 10^-8
[H+] =Ka*([H2PO4-]/[HPO4(2-)]
= (6.20 ×10^-8)×(0.155/0.1586)
= 6.059 ×10^-8 M
pH = - log[H+]
= - log (6.059×10^-8)
= 7.218
