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:

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

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

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.

quartz

explanation:

the density(ratio between mass and volume as m/v) decreases from top to bottom in periodic table group as the size increases and nuclear charge decreases. but in periods density increases from left to right as atom size decreases and nuclear charge increases.