Calcium reacts with water as shown. Ca(s) + 2H2O(l) → Ca(OH)2(aq) + H2(g) What is the total mass of the solution that remains when 40 g of calcium reacts with 100 g of water? A 58g B 74g C...

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Using the law of conservation of mass, the mass of the reactants should be equal to the mass of the products after the reaction. Hence, since we started with 100 grams of water and 40 grams of calcium, the total mass of reactants is 140 grams. Hence, we expect that at the end of the reaction, the total mass of the vessel is still 140 grams. However, we have hydrogen in the products side, which is a gas, and will hence evolve - thus decreasing the mass of the solution. We need to calculate the mass of hydrogen that evolves from the solution.

First, we need to find the limiting reagent. The balanced chemical equation is:

`Ca_((s)) +2H_2O_((l)) rarr Ca(OH)_(2(aq)) + H_(2(g))`

If 100 grams of water is used up:

`100 g H_2O * (1 mol H_2O)/(18 g H_2O) * (1 mol H_2)/(2 mol H_2O) * (2 g H_2)/(1 mol H_2) = 5.56 g H_2`

If 40 grams of calcium is used up:

`40 g Ca * (1 mol Ca)/(40 g Ca) * (1 mol H_2)/(1 mol Ca) * (2 g H_2)/(1 mol H_2) = 2 g H_2`

Hence, the limiting reagent is calcium, and the amound of hydrogen gas produced is 2 grams. This 2 grams of hydrogen gas will evolve, and thus the total mass of the solution will be decreased by 2 grams (since hydrogen gas escapes).

Hence, the mass that remains is 138 grams (letter C).




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