Answer:
m H2(g) = 2.241 g H2(g)
Explanation:
- 2Al(s) + 3H2SO4(aq) → Al2(SO4)3(aq) + 3 H2(g)
limit reagent:
∴ Mw Al = 26.982 g/mol
∴ Mw H2SO4 = 98.0785 g/mol
⇒ n Al = (20 g Al)×(mol/26.982 g) = 0.7412 mol Al
⇒ n H2SO4 = ( 115 g H2SO4 )×(mol/98.0785 g) = 1.173 mol H2SO4
⇒ n H2 = (0.7412 mol Al)×(3 mol H2/ 2 mol Al) = 1.112 mol H2
∴ Mw H2 = 2.016 g/mol
⇒ g H2 = (1.112 mol H2)×(2.016 g/mol) = 2.241 g H2
<span>c. remains the same
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As electrical energy is converted into heat energy, the total amount in the system remains the same
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according to law of conservation of energy
Answer:
18857.553
Explanation:
for an approximate result, multiply the value by 239
Answer:
mg3n2 and cacl3 hope it helps
