Answer:
53.7 grams of HNO3 will be produced
Explanation:
Step 1: Data given
Mass of NO2 = 59.0 grams
Molar mass NO2 = 46.0 g/mol
Step 2: The balanced equation
3NO2 + H2O→ 2HNO3 + NO
Step 3: Calculate moles NO2
Moles NO2 = 59.0 grams / 46.0 g/mol
Moles NO2 = 1.28 moles
Step 4: Calculate moles HNO3
For 3 moles NO2 we need 1 mol H2O to produce 2 moles HNO3 and 1 mol NO
For 1.28 moles NO2 we'll have 2/3 * 1.28 =0.853 moles HNO3
Step 7: Calculate mass HNO3
Mass HNO3 = 0.853 moles * 63.01 g/mol
Mass HNO3 = 53.7 grams
53.7 grams of HNO3 will be produced
Answer:
0.60 mol
Explanation:
There is some info missing. I think this is the original question.
<em>Acetylene gas is often used in welding torches because of the very high heat produced when it reacts with oxygen gas, producing carbon dioxide gas and water vapor. Calculate the moles of oxygen needed to produce 1.5 mol of water.</em>
<em />
Step 1: Given data
Moles of water required: 1.5 mol
Step 2: Write the balanced equation
C₂H₂(g) + 2.5 O₂(g) ⇒ 2 CO₂(g) + H₂O(g)
Step 3: Calculate the moles of oxygen needed to produce 1.5 mol of water
The molar ratio of O₂ to H₂O is 2.5:1. The moles of oxygen needed to produce 1.5 mol of water are (1/2.5) × 1.5 mol = 0.60 mol
Given that Ka = 4.3 × 10^-4
At the half equivalence (Stoichiometric) point;
pH = pKa = - log (Ka)
Therefore; pKa = - log (4.3 ×10^-4)
= 3.37
Therefore; the pH at the half-stoichiometric point is 3.37
