Mg(s) + 2HCl --> MgCl2 + H2...Limiting reactant?

Question

which is the limiting reactant...say you wre using 25 g of each, which would be limiting and why?

Answer 1

To determine limiting reagant you need a balanced equation (which you've already got) and the number of moles of reactants.

Divide the mass of each of your reactants (in grams) by it's molar mass (in grams) to get moles. Then from the reactant ratios of the balanced equation, you can tell which one runs out first. (usually, you can tell by looking at it)

For 25g of Mg: 25g/24.3g/mole = 1.03mole

for 25 g of HCl: 25/36.45g/mole = 0.686mole

From the equation we can see we need 2 HCl's for each Mg, and we already have less moles of it than the Mg, so it is the limiting reagant. Thus, the amount of MgCl2 that could be made is going to be 1/2 the number of moles of the HCl: or 0.686 (1/2) = 0.343mole.

If you need the mass of the product, just multiply moles of product by it's molar mass.

Answer 2

to find the limiting reactant, you need to take the moles of each substance and divide it by the coefficient of the balanced equation.

soo...

25g Mg x 1 mol / 24.3050 g = 1.0286 mol Mg
the coefficient of Mg is 1, and 1.0286/1 = 1.0286 mol Mg

25g HCl x 1 mol / 36.461 g = 0.6857 mol HCl
the coefficient of HCl is 2, so 0.6857/2 = 0.3428 mol HCl

the lowest number indicates the limiting reactant, so HCl would be your limiting reactant b/c it would run out first.

Answer 3

Which Is The Limiting Reactant

Answer 4

lol