If 1g of ba reacts with 1.8g al(so4)3, what is the excess reactant

Al₂(SO₄)₃ is the excess reactant.

Explanation:

Barium (Ba) react with Aluminium sulphate [Al₂(SO₄)₃] according to the following balanced equation:

3 Ba + Al₂(SO₄)₃ → 3 BaSO₄ + 2 Al

It is clear that 3 mol of Ba react with 1 mol of Al₂(SO₄)₃ to give 3 moles of BaSO₄

The limiting reactant is the reactant that produces the least amount of BaSO ₄.

The molar masses of each substance involved.

Ba : 137.3 g/mol

Al₂(SO₄)₃: 342.1 g/mol

BaSO ₄:  233.3 g/mol

Then we calculate no of moles of each reactant from the given mass.

As following:

no. of moles of Ba = (mass /molar mass) = (1 g / 137.3 g/mol) = 0.0073 mol

no. of moles of Al₂(SO₄)₃ = (mass /molar mass)

                                         = (1.8 g / 342.1 g/mol) = 0.0053 mol

Then we calculate mol of product produced from each reactant

For BaSO₄

3 mol of BaSO₄ produced from →  3 mol of Ba ?? mol of BaSO₄ produced from →  0.0073 mol of Ba

no of moles of BaSO₄ from Ba = (0.0073 * 3) / 3 = 0.0073 mol

then converting moles of BaSO₄ into mass

mass of BaSO₄ =  no of moles  * molar mass = 0.0073 * 233.3 = 1.7 g

For Al₂(SO₄)₃

3 mol of BaSO₄ produced from →  1 mol of Al₂(SO₄)₃ ?? mol of BaSO₄ produced from →  0.0053 mol of Al₂(SO₄)₃

no of moles of BaSO₄ from Al₂(SO₄)₃ = (0.0053 * 3) / 1 = 0.0159 mol

then converting moles of BaSO₄ into mass

mass of BaSO₄ =  no of moles  * molar mass = 0.0159 * 233.3 = 3.7 g

∴1 g of Ba  produces the least amount of barium sulfate, so it is the limiting reactant and Al₂(SO₄)₃ is the excess reactant.

For cyclohexane compounds drawn in a chair conformation:   -axial bonds are vertical  -equatorial bonds are parallel to two bonds within the ring.  thus, as i assume the chair conformation is drawn:   -the x group is in an axial position, as are the five other vertical bonds.  -three axial bonds are on the top face of the ring  -the other three axial bonds are on the bottom face  the position of the second substituent on each ring carbon is equatorial.  gauche groups must be on an adjacent carbon such as carbon 6 and carbon 2, and also gauche groups are rotated by 60 degrees relative to each other.  drawing the newman projections for c1–c2 and c1–c6 aids in visualizing the groups that are gauche to the y group on c1.  click this link and watch some academic videos that may you more understand the topic 

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