1.7 x 10 to the power of 22 formula units
A formula is the empirical formula of any ionic or covalent network solid compound used as an independent entity for stoichiometric calculations.
3.18 grams of product (Cu I) - 2.54 grams of Cu (reactant) = .64 grams of Sulfur, by law of conservation of mass.
Answer:
The correct answer is A. increasing the positive charge of the positively charged object and increasing the negative charge of the negatively charged object
I hope this helped :D
Answer is: <span>the percent yield for this reaction is 75,37%.
</span>m(MgCl₂) = 962 g.
n(MgCl₂) = m(MgCl₂) ÷ M(MgCl₂).
n(MgCl₂) = 962 g ÷ 95,21 g/mol.
m(MgCl₂) = 10,1 mol.
n(MgCl₂) : n(Mg) = 1 : 1.
n(Mg) = 10,1 mol.
m(Mg) = 10,1 mol · 24,3 g/mol.
m(Mg) = 245,43 g.
ω(Mg) = 185 g ÷ 245,43 g · 100% = 75,37%.
The equation for the reaction between NaOH and AlCl₃ is as follows;
3NaOH + AlCl₃ ---> 3NaCl + Al(OH)₃
the stoichiometry of NaOH : AlCl₃ is 3:1
3 moles of NaOH reacts with 1 mol of AlCl₃ to produce 1 mol of Al(OH)₃
the number of AlCl₃ moles reacted - 6.5 mol
molar mass of NaOH -(23 +16 +1) = 40 g/mol
the number of NaOH moles reacted = 57.0 g / 40 g/mol
NaOH moles = 1.425 mol
either NaOH or AlCl₃ is in excess and other is the limiting reactant.
limiting reactant is the reactant whose number of moles are fully consumed during the reaction. the reactant that is in excess will have leftover moles that are remaining after the reaction.
If AlCl₃ is the limiting reactant, number of NaOH moles would be thrice the amount of AlCl₃ present,
then number of NaOH moles that should be present - 6.5 * 3 = 19.5 mol
however there are only 1.425 mol of NaOH present, therefore AlCl₃ is in excess.
Then NaOH is the limiting reactant,
the amount of products formed depends on the amount of the limiting reactant present.
stoichiometry of NaOH : Al(OH)₃ is 3:1
the number of Al(OH)₃ moles produced = number of NaOH moles reacted / 3
number of Al(OH)₃ moles are - 1.425 mol /3 = 0.475 mol
molar mass of Al(OH)₃ = (27 +3*16 + 3*1) = 78 g/mol
mass of Al(OH)₃ produced = 78 g/mol * 0.475 mol = 37.05 g
