40 g of calcium (Ca) reacts with 0.4 M hydrochloric acid (HCl). The balanced equation for the reaction is:
`Ca + 2HCl -> H_2 + CaCl_2`
From the balanced reaction 1 mole of calcium requires 2 moles of pure hydrochloric acid.
The molar mass of calcium is approximately 40 g/mole. 40 g of calcium is equivalent to one mole of calcium. The complete chemical reaction of this with HCl requires two moles of hydrochloric acid.
0.4 M hydrochloric acid has 0.4 moles of the acid per liter of the solution.
2 moles HCl * (1 liter/0.4 moles) = 5 liters of solution
It requires 5 liters of 0.4 M HCl to react completely with 40 g of calcium.
Look at your balanced equation. It says for every ONE mole of calcium, you need TWO moles of hydrogen chloride to complete the reaction.
You've been given 40 grams of Ca so just convert that to moles first! First get the molar mass of Ca to convert the 40 g of Ca to moles.
Molar mass: 40.08 g/mol
40 g x (1 mol / 40.08 g ) = 1 mol
Because we have about 1 mol of Ca, we need 2 moles of HCl.
Plug in the 2 moles in your molarity equation.
Remember, M = mol/L
Plug in your molarity and mol values since you know them.
0.4 = 2/L
Solve for L and you get 5 liters.
In order for the reaction to occur, there needs to be twice the amount of moles HCl than of moles Ca.
First, solve for the number of moles of Ca.
`40"g" = (1"mol")/(40.08"g") = 1"mol"`
Therefore, you need 2 moles of HCl for the reaction to occur. Use the molarity equation:
`"M" = "mol"/"L"`
Plug in known values...
`0.4"M" = (2"mol")/"L"`
... and solve for the unknown variable.
`"L" = 5`
You would need 5 L. Looking at the problem and the sig figs, you are only required to have one sig fig!