Answer:
14.336 g MnF₂
Explanation:
number of moles = mass / molecular weight
number of moles of MnI₂ = 55 / 309 = 0.178 moles
number of moles of F₂ = 55 / 38 = 1.447 moles
From the reaction and the number of moles calculated we deduce that the fluorine F₂ is a limiting reactant.
So:
if 13 moles of F₂ reacts to produce 2 moles of MnF₃
then 1.447 moles of F₂ reacts to produce X moles of MnF₃
X = (1.447 × 2) / 13 = 0.223 moles of MnF₃ (100% yield)
For 57.2% yield we have:
number of moles of MnF₃ = (57.2 / 100) × 0.223 = 0.128 moles
mass = number of moles × molecular weight
mass of MnF₃ = 0.128 × 112 = 14.336 g
The reaction between oxygen, O2, and hydrogen, H2, to produce water can be expressed as,
2H2 + O2 --> 2H2O
The masses of each of the reactants are calculated below.
2H2 = 4(1.01 g) = 4.04 g
O2 = 2(16 g) = 32 g
Given 1.22 grams of oxygen, we determine the mass of hydrogen needed.
(1.22 g O2)(4.04 g H2 / 32 g O2) = 0.154 g of O2
Since there are 1.05 grams of O2 then, the limiting reactant is 1.22 grams of oxygen.
<em>Answer: 1.22 g of oxygen</em>
When a neutral hydrogen atom loses an electron, and positively-charged particles should remain. A proton should be 1840 times heavier than an electrons. Cathode rays should be attracted to a positively- charged plate.
Carbon-14, 14C, or radiocarbon, is a radioactive isotope of carbon with a nucleus containing 6 protons and 8 neutrons.
You need to know that the atomic number of nitrogen is 7, and you subtract that from the mass number to get16-7= 9.
Answer:
630.95 grams of Na₂CO₃ would be needed to produce 1000g of NaHCO₃
Explanation:
The balanced reaction is:
Na₂CO₃ + CO₂+ H₂O → 2 NaHCO₃
By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of each compound participate in the reaction:
- Na₂CO₃: 1 moles
- CO₂: 1 mole
- H₂O: 1 mole
- NaHCO₃: 2 moles
Being the molar mass:
- Na₂CO₃: 106 g/mole
- CO₂: 44 g/mole
- H₂O: 18 g/mole
- NaHCO₃: 84 g/mole
Then by stoichiometry the following quantities of mass participate in the reaction:
- Na₂CO₃: 1 mole* 106 g/mole= 106 g
- CO₂: 1 mole* 44 g/mole= 44 g
- H₂O: 1 mole* 18 g/mole= 18 g
- NaHCO₃: 2 moles* 84 g/mole= 168 g
You can apply the following rule of three: if 106 grams of Na₂CO₃ are needed to produce 168 grams of NaHCO₃, how much mass of Na₂CO₃ is necessary to produce 1000 grams of NaHCO₃?
mass of Na₂CO₃= 630.95 grams
<u><em>630.95 grams of Na₂CO₃ would be needed to produce 1000g of NaHCO₃</em></u>
