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
C9H20 + 14O2 --> 9CO2 + 10H2O
<h2>Step 1 : Identify the given </h2>
Volume = 250mL
Density = 1.19 g/ML
<h2>Step 2 . Calculate the mass of HCL </h2>
Density = mass/volume
∴Mass = Density * Volume
= 1.19g/mL* 250mL
= 297,5g
<h2>Step 3 : Calculate the total mass of the solution, given that concentration HCL is 38% </h2>
Mass of the total solution can be calculated by the following :
38% = Mc /297.5 * 100
Mc = 38/100 *297.5
= 113.05grams
• Finally, this means that mass of the total solution of 0.125M HCL i,s 113grams, ,you would use this mass to prepare 250 mL of 0.125 M HCl from concentrated HCl (aq) that is 38.0%
If you are asking for the word for this definition it is a <span>attraction by the two nucluei</span>
There are 1.078 x 10²³ molecules
<h3>Further explanation</h3>
Given
4 dm³ = 4 L Nitrogen gas
Required
Number of molecules
Solution
Assumptions on STP (1 atm, 273 K), 1 mol gas = 22.4 L, so for 4 L :
mol = 4 : 22.4
mol = 0.179
1 mol = 6.02 x 10²³ particles(molecules, atoms)
For 0.179 :
= 0.179 x 6.02 x 10²³
= 1.078 x 10²³
