Question

Consider this balanced chemical equation: Ca + 2H₂O    -> Ca(OH)₂ + H₂

The limiting reactant when 3.00 moles of calcium are reacted with 8.00 moles of water in the above equation is ____________________.

If you carried out the above reaction with these molar amounts, what will be the theoretical yield of hydrogen gas. _______________________g.

A student carried out the reaction above and collected 2.15 g of hydrogen gas product. What is the % yield for the reaction? _______________________. (3 sig figs)

Answer 1

Given

Ca + 2H2O ==> Ca(OH)2 + H2

From the equation we can say that

1 mole of Ca reacts with 2 moles of H2O

But given 3 moles of Ca

So

3 moles of Ca reacts with ( 3 moles of Ca * 2 moles of H2O/1 mole of Ca)

= 6 moles of H2O

But there are 8 moles of H2O

Thus Ca is the limiting reactant

____________________

From the reaction we can say that 1 mole of Ca produces 1 mole of Ca(OH)2 and 1 mole of H2

Hence the products formed will be 3 moles of Ca(OH)2 and 3 moles of H2 ( as Ca is limiting reactant it will be considered for formation of products)

Thus 3 moles of H2 is formed

Mass of H2 = moles of H2 * molar mass of H2

= (3 moles * 2 g/mole)

= 6 grams

so 6 grams is the theoretical yield of Hydrogen gas

____________________

Given

2.15 g of hydrogen gas is formed

% yield = (actual yield / theoretical yield) * 100

= (2.15 g / 6 g) * 100

= 35.8 %

So

35.8 % is the % yield