Calculate the mass of water and salt that are formed during the neutralization of 98g. Ca (OH) 2 with HNO3

Given:

m (Ca (OH) 2) = 98 g;

Find:

m (Ca (NO3) 2) -?

m (H2O) -?

Solution:

1) We compose the reaction equation typical for the problem condition:

Ca (OH) 2 + 2HNO3 = H2O + Ca (NO3) 2;

2) Find the amount of calcium hydroxide contained in 100 grams of salt:

n (Ca (NO3) 2) = m: M = 98 g: 100 g / mol = 0.98 mol;

3) We find the amount and mass of water according to the first logical equality:

if 1 mol of Ca (OH) 2 gives 1 mol of H2O,

then 0.98 mol Ca (OH) 2 will give x mol H2O,

then x = 0.98 mol.

m (H2O) = n * M = 0.98 mol * 18 g / mol = 17.64 g;

4) Find the amount and mass of calcium nitrate by the second logical expression:

if 1 mol of Ca (OH) 2 gives 1 mol of Ca (NO3) 2,

then 0.98 mol Ca (OH) 2 will give x mol Ca (NO3) 2,

then x = 0.98 mol.

m (Ca (NO3) 2) = n * M = 0.98 mol * 164 g / mol = 160.72 g;

Answer: m (H2O) = 17.64 g; m (Ca (NO3) 2) = 160.72 g.