Answer:
About 16.1 grams of oxygen gas.
Explanation:
The reaction between magnesium and oxygen can be described by the equation:
24.4 grams of Mg reacted with O₂ to produce 40.5 grams of MgO. We want to determine the mass of O₂ in the chemical change.
Compute using stoichiometry. From the equation, we know that two moles of MgO is produced from every one mole of O₂. Therefore, we can:
- Convert grams of MgO to moles of MgO.
- Moles of MgO to moles of O₂
- And moles of O₂ to grams of O₂.
The molecular weights of MgO and O₂ are 40.31 g/mol and 32.00 g/mol, respectively.
Dimensional analysis:
In conclusion, about 16.1 grams of oxygen gas was reacted.
You will obtain the same result if you compute with the 24.4 grams of Mg instead:
Particle accelerators is your answer
Answer:
Darmstadtium
Explanation:
An element with the electronic configuration 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p⁶6s²4f¹⁴5d¹⁰6p⁶7s²5f¹⁴6d⁸ has 110 electrons in its electron shells.
Since the element is a neutral atom, this number is also equal to its atomic number. Therefore, its atomic number is 110.
The element in the period table that has an atomic number of 110 is Darmstadtium, a d-block element, thus a transittion metal. It also belong to period 7 in the Periodic table of elements.
Answers:
<span>Answer 1: 10.03 g of siver metal can be formed.</span>
Answer 2: 3.11 g of Co are left over.
Work:
1) Unbalanced chemical equation (given):
<span>Co + AgNO3 → Co(NO3)2 + Ag
2) Balanced chemical equation
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<span>Co + 2AgNO3 → Co(NO3)2 + 2Ag
3) mole ratios
1 mol Co : 2 mole AgNO3 : 1 mol Co(NO3)2 : 2 mol Ag
4) Convert the masses in grams of the reactants into number of moles
4.1) 5.85 grams of Co
# moles = mass in grams / atomic mass
atomic mass of Co = 58.933 g/mol
# moles Co = 5.85 g / 58.933 g/mol = 0.0993 mol
4.2) 15.8 grams of Ag(NO3)
# moles Ag(NO3) = mass in grams / molar mass
molar mass AgNO3 = 169.87 g/mol
# moles Ag(NO3) = 15.8 g / 169.87 g/mol = 0.0930 mol
5) Limiting reactant
Given the mole ratio 1 mol Co : 2 mol Ag(NO3) you can conclude that there is not enough Ag(NO3) to make all the Co react.
That means that Ag(NO3) is the limiting reactant, which means that it will be consumed completely, whilce Co is the excess reactant.
6) Product formed.
Use this proportion:
2 mol Ag(NO3) 0.0930mol Ag(NO3)
--------------------- = ---------------------------
2 mol Ag x
=> x = 0.0930 mol
Convert 0.0930 mol Ag to grams:
mass Ag = # moles * atomic mass = 0.0930 mol * 107.868 g/mol = 10.03 g
Answer 1: 10.03 g of siver metal can be formed.
6) Excess reactant left over
1 mol Co x
----------------------- = ----------------------------
2 mole Ag(NO3) 0.0930 mol Ag(NO3)
=> x = 0.0930 / 2 mol Co = 0.0465 mol Co reacted
Excess = 0.0993 mol - 0.0465 mol = 0.0528 mol
Convert to grams:
0.0528 mol * 58.933 g/mol = 3.11 g
Answer 2: 3.11 g of Co are left over.
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