how many grams of H2 woulb be produce if 130g of zinc reacts with an excess dilute H2So4(Zn-65)?
Answers
Answer:
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Explanation:
Zn(s)+2HCl(aq)→ZnCl
2
(aq)+H
2
(g)
1 mole of Zn reacts with 2 moles of HCl to produce 1 mole of hydrogen.
The atomic mass of Zn is 65.4 g/mol. The molar mass of HCl is 36.5 g/mol.
When we divide mass with atomic/molar mass, we get number of moles.
Number of moles of Zn =
65.4
65
=0.994 moles.
Number of moles of HCl =
36.5
65
=1.78 moles.
0.994 moles of Zn will react with 2×0.994=1.988 moles of HCl however only 1.78 moles of HCl are present. Hence, HCl is the limiting reactant.
1.78 moles of HCl will produce
2
1.78
=0.89 moles of H
2
.
The molar mass of H
2
is 2 g/mol.
Mass of H
2
obtained =2g/mol×0.89mol=1.78g=1.8g
Answer:
Zn(s)+2HCl(aq)→ZnCl
2
(aq)+H
2
(g)
1 mole of Zn reacts with 2 moles of HCl to produce 1 mole of hydrogen.
The atomic mass of Zn is 65.4 g/mol. The molar mass of HCl is 36.5 g/mol.
When we divide mass with atomic/molar mass, we get number of moles.
Number of moles of Zn =
65.4
65
=0.994 moles.
Number of moles of HCl =
36.5
65
=1.78 moles.
0.994 moles of Zn will react with 2×0.994=1.988 moles of HCl however only 1.78 moles of HCl are present. Hence, HCl is the limiting reactant.
1.78 moles of HCl will produce
2
1.78
=0.89 moles of H
2
.
The molar mass of H
2
is 2 g/mol.
Mass of H
2
obtained =2g/mol×0.89mol=1.78g=1.8g