Answer:
2 mol of CO₂
Solution:
The reaction is as follow,
H₂CO + O₂ → CO₂ + H₂O
According to this equation,
1 mole of H₂CO produces = 1 mole of CO₂
So,
2 moles of H₂CO will produce = X moles of CO₂
Solving for X,
X = (2 mol × 1 mol) ÷ 1 mol
X = 2 mol of CO₂
Answer:
The claim is untrue.
Explanation:
Given the information from the question. We need to evaluate the claim.
According to the phase rule we have F= C-P+2
In this particular situation, the forms are completely allotropic .In order words, they are conjured through the same chemical composition. Thus constituents C= 1, P=4 for four phases and the number variables is 2. As a result, F= C-P+2 =1-4+2= -1. Therefore, the claim is untrue.
Hello friends..
find the MW of HP
calculate the # of mols per 100g of HP
take the # of mols times 6.023 x 10^23 times 2 (2 H per molecule)
Hope it helps you..
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
Each orbital must contain a single electron before any orbital contains two electrons.
