Im going to say that it is 1.) The Heart
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
As per Ideal gas equation, molar mass of the gas is 5.032 g/mo
We’ll begin by calculating the number of mole of the gas. This can be obtained as follow:
Volume (V) = 1.6 L
Temperature (T) = 287 K
Pressure (P) = 0.92 atm
Gas constant (R) = 0.0821 atm.L/Kmol
Number of mole (n) =?
According to Ideal gas equation , PV = nRT
0.92 × 1.6 = n × 0.0821 × 287
1.472 = n × 23.5627
Divide by 23.5627
n = 1.176 / 23.5627
n = 0.0624 mole
Finally, we shall determine the molar mass of the gas. This can be obtained as follow:
Mass of gas = 0.314 g
Number of mole = 0.0624 mole
Mole = 
0.0624 = 
Cross multiply
0.0624 × molar mass of gas = 0.314
Divide by 0.0624
Molar mass of gas = 
Molar mass of gas = 5.032 g/mo
Therefore the Molar mass of gas is 5.032 g/mo
Learn more about Ideal gas equation here:
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<h3><u>Answer;</u></h3>
= 9.45 × 10^23 molecules
<h3><u>Explanation; </u></h3>
The molar mass of Na2SO4 = 142.04 g/mol
Number of moles = mass/molar mass
= 223/142.04
= 1.57 moles
But;
1 mole = 6.02 × 10^23 molecules
Therefore;
1.57 moles = ?
= 1.57 × 6.02 × 10^23 molecules
<u>= 9.45 × 10^23 molecules </u>
