Answer:
7.82x10^24 molecules of water
Explanation:
H2O=18.015 g/mol Avogadro's #=6.022x10^23 molecules
0.234L x 1000g/1L x 1 mol H2O/18.015 g x 6.022x10^23 = 7.82x10^23 molecules of water
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 formula is SO2F2
Explanation:
Step 1: Data given
Suppose the mass of compound = 100 grams
The compound contains:
31.42 % S = 31.42 grams S
31.35 % O = 31.35 grams O
100 - 31.42 - 31.35 = 37.23 F
Molar mass of S = 32.065 g/mol
Molar mass F = 19.00 g/mol
Molar mass O = 16.00 g/mol
Step 2: Calculate moles
Moles = mass / molar mass
Moles S = 31.42 grams / 32.065 g/mol
Moles S = 0.9799 moles
Moles 0 = 31.35 grams / 16.00 g/mol
Moles 0 = 1.959 moles
Moles F = 37.23 grams / 19.00 g/mol
Moles F = 1.959 moles
Step 3: Calculate mol ratio
We divide by the smallest amount of moles
S: 0.9799 / 0.9799 = 1
F: 1.959/ 0.9799 = 2
O : 1.959 / 0.9799 = 2
The empirical formula is SO2F2
This formula has a molecular mass of 102.06 g/mol
This means the empirical formula is also the molecular formula : SO2F2
The mass (in grams) of iron, Fe that can be made from 21.5 g of Fe₂O₃ is 15.04 g
We'll begin by writing the balanced equation for the reaction. This is given below:
2Fe₂O₃ -> 4Fe + 3O₂
- Molar mass of Fe₂O₃ = 159.7 g/mol
- Mass of Fe₂O₃ from the balanced equation = 2 × 159.7 = 319.4 g
- Molar mass of Fe = 55.85 g/mol
- Mass of Fe from the balanced equation = 4 × 55.85 = 223.4 g
From the balanced equation above,
319.4 g of Fe₂O₃ decomposed to produce 223.4 g of Fe
<h3>How to determine the mass of iron, Fe produced</h3>
From the balanced equation above,
319.4 g of Fe₂O₃ decomposed to produce 223.4 g of Fe
Therefore,
21.5 g of Fe₂O₃ will decompose to produce = (21.5 × 223.4) / 319.4 = 15.04 g of Fe
Thus, 15.04 g of Fe were produced.
Learn more about stoichiometry:
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