Zn + 2HCl ---> ZnCl2 + H2(g)
What mass of zinc is needed to produce 115 mL of hydrogen?
t's given 115 mL of hydrogen, so can convert to 115/1000/22,4 = 0.0051 moles H2
- the ratio between HCl and H2 is 2 : 1, so the moles of HCl needed for this reaction is 0.01
- the ratio between Zn and H2 is 1 : 1, so the moles of Zn needed for this reaction is 0.0051
so the mass of Zn need is 0.0051*65 = 0.3315 (g)
How many moles of HCl are needed for the reaction? (5 marks)
confused by moles part
moles HCl needed = 2 x 0.0051 since stoichiometric ratio is 2:1
moles HCl = 0.01 moles
You are correct in saying that
the ratio between HCl and H2 is 2 : 1, so the moles of HCl needed for this reaction is 0.01
yes if at Standard temp and pressure, 22.4 liters/mol
so
0.115 liters / 22.4 = 0.00513 mols of H2 and of Zn
so yes 0.00513 mols * 65.4 grams/mol = 0.336 grams of zinc
close enough we agree
Now for every mol of H2 you need 2 mols of HCl so we agree again
2 * 0.00513 = 0.01 mols HCl
this curriculum has been used for a few years and all the answers are online XD
yup lol i've seen some of the exact same questions online from posts that are 9 yrs ago, think it goes further back if I recall based on some others haha.
To determine the number of moles of HCl needed for the reaction, we need to use the balanced chemical equation:
Zn + 2HCl -> ZnCl2 + H2
From the equation, we can see that the ratio between HCl and Zn is 2:1. This means that for every 2 moles of HCl, we need 1 mole of Zn.
Since we know the moles of Zn needed is 0.0051 moles (as calculated in the previous question), we can use this information to find the moles of HCl needed.
Using the ratio of 2 moles of HCl to 1 mole of Zn, we can set up a proportion:
2 moles HCl / 1 mole Zn = X moles HCl / 0.0051 moles Zn
Cross-multiplying and solving for X, we get:
2 moles HCl * 0.0051 moles Zn = X moles HCl * 1 mole Zn
X = (2 moles HCl * 0.0051 moles Zn) / 1 mole Zn
X = 0.0102 moles HCl
Therefore, the number of moles of HCl needed for the reaction is 0.0102 moles.