Na2SO3 + 2 AlCl3 --> Al2(SO3)3 + 6 NaCl Using the above reaction, suppose (1.760x10^2) grams of Na2SO3 were added to (6.5x10^1) grams of AlCl3, how much (in g) of NaCl was produced once the reaction was run to completion?

Mass NaCl produced : 171.81 g

The reaction equation is the chemical formula of reagents and product substances  

A reaction coefficient is a number in the chemical formula of a substance involved in the reaction equation. The reaction coefficient is useful for equalizing reagents and products  

Na₂SO₃ : 176 g

MW : 126,043 g/mol

mol :

AlCl₃ : 65 g

MW : 133,34 g/mol

mol :

Reaction

Na₂SO₃ + 2 AlCl₃ ⇒ Al₂(SO₃)₃ + 6NaCl

limiting reactants:

mol ratio(mol/coefficient) Na₂SO₃ : 2 AlCl₃ =

AlCl₃ ⇒ limiting reactants(smaller ratio)

mol NaCl from limiting reactants :

mass NaCl

2b2t

explanation:

2b2t

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2