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
Full Question:
Ammonia chemically reacts with oxygen gas to produce nitric oxide and water. What mass of water is produced by the reaction of 7.7g of ammonia?
Be sure your answer has the correct number of significant digits.
Answer:
12.23g ≈ 12g (2 s.f)
Explanation:
Ammonia chemically reacts with oxygen gas to produce nitric oxide and water. The balanced chemical reaction is given as:
4 NH3 + 5 O2 -------> 4 NO + 6 H2O
From the reaction;
4 mole of ammonia reacts to produce 6 moles of water
From the question;
Moles = mass / molar mass
From the question;
moles of ammonia = mass / molar mass = 7.7 / 17 = 0.4529moles
Number of moles of water produced;
4 = 6
0.4529 = x
x = (0.4529 * 6 ) / 4
x = 0.67935moles
Mass of water = moles * molar mass = 0.67935 * 18 = 12.23g ≈ 12g (2 s.f)
Option B. <span>Rb2O + Cu(C2H3O2)2 → 2RbC2H3O2 + CuO is the correct answer. HOPE IT HELPS</span>
