Answer:
We will expect 4 moles of MgO to be formed (option b).
Explanation:
Step 1: The balanced equation
2Mg + O2 → 2MgO
Step 2: Data given
Number of moles of Magnesium = 4 moles
Oxygen = in excess → this means Magnesium is the limiting reactant
Magnesium will completely be consumed ( 4 moles). There will remain 0 moles.
For 2 moles of magnesium consumed, we need 1 mole of oxygen to produce 2 moles of MgO.
For 4 moles of magnesium, we need 4/2 = 2 moles of oxygen.
For 4 moles of magnesium, we will produce 4/1 = 4 moles of MgO
We will expect 4 moles of MgO to be formed (option b).
Answer:
2,75 mol of O2 it's 88 g of O2.
Explanation:
The weight of the diatomic molecule O2 is 32 g/mol. So considering that, you should multiply 2,75 mol · 32 = 88g :)
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: D=8.27 g/cm³
Explanation:
Density is mass/volume. Mass is in grams and volume is in liters. In this case, the problem wants our volume to be in cm³. All we need to do is to make some conversions to convert kg/m³ to g/cm³.

With this equation, the m³ and kg cancel out, and we are left with g/cm³.
D=8.27 g/cm³
15 grams of NH3 can be dissolved
<h3>Further explanation</h3>
Given
50 grams of water at 50°C
Required
mass of NH3
Solution
Solubility is the maximum amount of a substance that can dissolve in some solvents. Factors that affect solubility
- 1. Temperature:
- 2. Surface area:
- 3. Solvent type:
- 4. Stirring process:
We can use solubility chart (attached) to determine the solubility of NH3 at 50°C
From the graph, we can see that the solubility of NH3 in 100 g of water at 50 C is 30 g
So that the solubility in 50 grams of water is:
= 50/100 x 30
= 15 grams