limiting reagents , help!?

moles of HCl = MV (L) = 2 x 20 / 1000 = 0.04 moles

Molar mass of Na2CO3 is 106.0 g/mol

moles of Na2CO3 = 5 / 106 = 0.0472 moles

from the balanced equation we see that 1 mole of Na2CO3 r/w 2 mole of HCl

0.0472 moles of Na2CO3 needs 0.0943 of HCl for a complete reaction but we only have .04 moles of HCl so

HCl is the limiting reagent and forms CO2 = 1/2 x moles of HCl

= .02 moles or .02 x 44 = .88 g of CO2

A proscribing reagent is only the reactant that runs out first for the time of a reaction. As a reaction proceeds, reactants cut back, at the same time as products boost. The proscribing reagent is only the reactant it extremely is used up first. An extra reactant is only the different reactant left over after the proscribing reagent is used. there is often a proscribing reagent, it only relies upon on what share moles of what chemical you have.

To find the amount of CO2 produced

M= mol solute/ L solution

(0.02 L)(2 mol HCL/L solution)(1 mol CO2 / 2 mol HCL) =0.02 mol CO2

Na2CO3= 105.99 g/mol

(5 g Na2CO3)(1 mol Na2CO3 / 105.99 g Na2CO3)(1 mol CO2/ 1 mol Na2CO3) = 0.04 mol CO2

Since HCL can only produce 0.02 mol CO2 which is less than the amount by Na2CO3 the answer is HCL is the limiting reagent and 0.02 mol CO2 or to convert

(0.02 mol CO2)(44.01 g CO2 / mol CO2)= 0.88 g CO2

Volume HCl = 20 ml = 0.020L ( remember your sig figs)

Molarity HCl = 2.0M = 2.0 mol.L-1

Number of mols= molarity x volume

=2.0 mol.L-1 X 0.020L

=0.040 mol HCl

Ratio HCl: CO2

2 : 1

mols CO2 = MOLS HCl/2

= 0.040/2 =0.02 mol CO2

mass= n x M

= 0.02MOL X (12.01+ 2(15.99))

= 0.88G CO2

mol Na2CO3 = mass/molarmass

=5/ 2(22.99) +12.0 +3(15.99)

=0.047 mol

therefore number of mols of HCl is less than that of Na2CO3

therefore HCl is the limiting reagent

When given molarity and volume, you can work out the number of moles used by multiplyin them, since M=n/V---> MV=n. So, 2.0M HCL, 0.02 liters of that, contains 0.04 moles of HCl. By stoichiometry: (0.04 moles HCl)(1 mole CO2/2 moles HCl)=0.02 moles CO2 produced IF HCL IS THE LIMITING REAGENT.

Given the mass of a substance, you can get the number of moles using its molar mass; moles=mass/molar mass. So,( 5gNa2CO3)/(105.99gNa2CO3/mole) gives you 0.047 moles Na2CO3.

By stoichiometry: (0.47 moles Na2CO3)(1 mol CO2/1 mole Na2CO3)= 0.047 moles CO2 produced, IF NA2CO3 IS THE LIMITING REACTANT.

Since HCl produced less moles of CO2, it is the limiting reactant in this reaction.