The sample must be sufficiently soluble (fig. 2) to yield an NMR spectrum. For 1H and 1H observed NMR, it is recommended to dissolve between 2 and 10 mg in between 0.6 and 1 mL of solvent so that the sample depth is at least 4.5 cm in the tube (fig. 3).
Answer:
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Answer:
a. NH3 is limiting reactant.
b. 44g of NO
c. 40g of H2O
Explanation:
Based on the reaction:
4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(l)
4 moles of ammonia reacts with 5 moles of oxygen to produces 4 moles of NO and 6 moles of water.
To find limiting reactant we need to find the moles of each reactant and using the balanced equation find which reactant will be ended first. Then, with limiting reactant we can find the moles of each reactant and its mass:
<em>a. </em><em>Moles NH3 -Molar mass. 17.031g/mol-</em>
25g NH3*(1mol/17.031g) = 1.47moles NH3
Moles O2 = 4 moles
For a complete reaction of 4 moles of O2 are required:
4mol O2 * (4mol NH3 / 5mol O2) = 3.2 moles of NH3.
As there are just 1.47 moles, NH3 is limiting reactant
b. Moles NO:
1.47moles NH3 * (4mol NO/4mol NH3) = 1.47mol NO
Mass NO -Molar mass: 30.01g/mol-
1.47mol NO * (30.01g/mol) = 44g of NO
c. Moles H2O:
1.47moles NH3 * (6mol H2O/4mol NH3) = 2.205mol H2O
Mass H2O -Molar mass: 18.01g/mol-
2.205mol H2O * (18.01g/mol) = 40g of H2O
Answer : The molecular weight of a gas is, 128.9 g/mole
Explanation : Given,
Density of a gas = 5.75 g/L
First we have to calculate the moles of gas.
At STP,
As, 22.4 liter volume of gas present in 1 mole of gas
So, 1 liter volume of gas present in 
mole of gas
Now we have to calculate the molecular weight of a gas.
Formula used :
Now put all the given values in this formula, we get the molecular weight of a gas.
Therefore, the molecular weight of a gas is, 128.9 g/mole
