2.5 mol Zn and 8.75 mol HCl react according
to the equation
Zn + 2 HCl −→ ZnCl2 + H2 .
Calculate the amount in moles of ZnCl2
formed.
Answer in units of mol
AgNO3 + NaCl → AgCl + NaNO3 .
How many grams of NaCl would be required
to react with 446 mL of 0.211 M AgNO3
solution?
Answer in units of grams
4 answers
mols AgNO3 = M x L = ?
Using the coefficients in the balanced equation, convert mols AgNO3 to mols NaCl.
Now convert mols NaCl to grams. g = mols x molar mass.
Using the coefficients in the balanced equation, convert mols AgNO3 to mols NaCl.
Now convert mols NaCl to grams. g = mols x molar mass.
Your second problem is a limiting reagent (LR) problem. You know it is a LR problem because amounts are given for BOTH reactants. You do these just like simple stoichhiometry problem BUT do it for EACH reagent then take the smaller amount formed as the product formed.
Convert mols Zn to mols ZnCl2.
2.5 mol Zn will give you 2.5 mol ZnCl2 (1:1 ratio in the equation) if you had HCl in excess.
Do the same for HCl. 8.75 mols HCl will form 1/2(8.75) = 4.375 mols ZnCl2 if you had Zn in excess.
The smaller number is 2.5; therefore, 2.5 mols Zn will be formed, Zn is the limiting reagent, and HCl will be in excess which means some of it will react and there will be some left over.
Then convert 2.5 mol Zn to grams. g = mols x molar mass.
Convert mols Zn to mols ZnCl2.
2.5 mol Zn will give you 2.5 mol ZnCl2 (1:1 ratio in the equation) if you had HCl in excess.
Do the same for HCl. 8.75 mols HCl will form 1/2(8.75) = 4.375 mols ZnCl2 if you had Zn in excess.
The smaller number is 2.5; therefore, 2.5 mols Zn will be formed, Zn is the limiting reagent, and HCl will be in excess which means some of it will react and there will be some left over.
Then convert 2.5 mol Zn to grams. g = mols x molar mass.
How many grams of NaCl would be required
to react with 594 mL of 0.278 M AgNO3
solution?
Answer in units of grams.
to react with 594 mL of 0.278 M AgNO3
solution?
Answer in units of grams.
1 answer