First, let us write the reaction. We know that the reactants are Aluminum and Sulfuric acid, and one of the products is hydrogen. That means the reaction is a single replacement reaction as shown below:
2 Al + 3 H₂SO₄ = 3 H₂ + Al₂(SO₄)₃
Next, let us determine which of the reactants is limiting:
1.80 g Al (1 mol/26.98 g)(3 mol H₂SO₄/2 mol Al)(98 g H₂SO₄/mol) = 9.807 g
It would need 9.807 g. But the only available amount is 6 g. That means that H₂SO₄ is the limiting reactant. We use 6 g as the basis to know the theoretical yield:
6 g*(1 mol H₂SO₄/98 g)*(3 mol H₂/ 3 mol H₂SO₄)*(2 g /mol H₂) = 0.122 g
Therefore, the percent yield is equal to
Percent yield = (0.112 g/0.122 g)*100
Percent yield = 91.8%
Molarity = moles / Volume in liters
M = 0.45 / 3.0
M = 0.15 mol/L
Answer:
56765
Explanation:
i did the math and this is the answer
4, 3, 1
Explanation:
Sb has four on the right, so it needs four on the left. It's all alone, so <u>4</u>.
O2 comes in pairs, so you only need <u>3</u> of those to get six oxygens.
And 1 is the place holder for the right side since we got the left to match perfectly.
