Asked by samrawit

If 6.5g of zinc react with 5.0 of HCl according to the following reactant, which substance is the limiting reactant? How many grams of the reactant remain unreacted? How many grams of hydrogen would be produced?
Answered by DrBob222
5.0 WHAT of HCl. I assume 5.0 grams.
Zn + 2HCl ==> ZnCl2 + H2
mols Zn = 6.5/65.4= 0.099
mols HCl = 5.0/36.5 = 0.188
mols H2 produced with Zn and excess HCl = 0.099 mol Zn x (1 mol H2/1 mol Zn) = 0.099 mol H2.
mols HCl produced with HCl and excess Zn = 0.188 x (1 mol H2/2 mol HCl) = 0.188 x 1/2 = 0.094
In limiting reagent (LR) problems the smaller number wins since you can't produce more than the smaller amount so HCl is the LR and Zn is the excess reagent (ER).
Grams H2 produced = 0.094 x molar mass H2 = ?
All of the HCl is used. There is no HCl left unreacted.
How much Zn is used in the reaction? That 0.188 mols HCl x (1 mol Zn/2 mol HCl) = 0.094. You had 0.099 initially. Zn unreacted is 0.099-0.094 = ?. Convert that to grams by g Zn = mols Zn x 65.4 = ?

Answered by Rediet
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Answered by Seid
Make it short
Answered by muazu
what is the formular of volume
Answered by Gemechis asrat
5.0 WHAT of HCl. I assume 5.0 grams.
Zn + 2HCl ==> ZnCl2 + H2
mols Zn = 6.5/65.4= 0.099
mols HCl = 5.0/36.5 = 0.188
mols H2 produced with Zn and excess HCl = 0.099 mol Zn x (1 mol H2/1 mol Zn) = 0.099 mol H2.
mols HCl produced with HCl and excess Zn = 0.188 x (1 mol H2/2 mol HCl) = 0.188 x 1/2 = 0.094
In limiting reagent (LR) problems the smaller number wins since you can't produce more than the smaller amount so HCl is the LR and Zn is the excess reagent (ER).
Grams H2 produced = 0.094 x molar mass H2 = ?
All of the HCl is used. There is no HCl left unreacted.
How much Zn is used in the reaction? That 0.188 mols HCl x (1 mol Zn/2 mol HCl) = 0.094. You had 0.099 initially. Zn unreacted is 0.099-0.094 = ?. Convert that to grams by g Zn = mols Zn x 65.4 = ?
Answered by B
HCl is limiting reactant
7.1g remains uncreated
1.8g of hydrogen produced
Answered by merwa mohammed
5.0 WHAT of HCl. I assume 5.0 grams.
Zn + 2HCl ==> ZnCl2 + H2
mols Zn = 6.5/65.4= 0.099
mols HCl = 5.0/36.5 = 0.188
mols H2 produced with Zn and excess HCl = 0.099 mol Zn x (1 mol H2/1 mol Zn) = 0.099 mol H2.
mols HCl produced with HCl and excess Zn = 0.188 x (1 mol H2/2 mol HCl) = 0.188 x 1/2 = 0.094
In limiting reagent (LR) problems the smaller number wins since you can't produce more than the smaller amount so HCl is the LR and Zn is the excess reagent (ER).
Grams H2 produced = 0.094 x molar mass H2 = ?
All of the HCl is used. There is no HCl left unreacted.
How much Zn is used in the reaction? That 0.188 mols HCl x (1 mol Zn/2 mol HCl) = 0.094. You had 0.099 initially. Zn unreacted is 0.099-0.094 = ?. Convert that to grams by g Zn = mols Zn x 65.4 = ?
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