# In 120 ml of a solution of nitric acid with a mass fraction of 7% (density 1.03 g / ml) made 12.8 g of calcium carbide

**In 120 ml of a solution of nitric acid with a mass fraction of 7% (density 1.03 g / ml) made 12.8 g of calcium carbide. How many milliliters of 20% hydrochloric acid (density 1.10 g / ml) should be added to the resulting mixture to completely neutralize it?**

Find the mass of the HNO3 solution.

m = Vp.

m = 1.03 g / ml × 120 ml = 8.65 g.

Let’s find the amount of the substance HNO3.

M (HNO3) = 63 g / mol.

n = 8.65 g: 63 g / mol = 0.14 mol.

Let’s find the amount of substance CaC2.

M (CaC2) = 64 g / mol.

n = 12.8 g: 64 g / mol = 0.2 mol (excess).

Let’s find the quantitative ratios of substances.

2HNO3 + CaC2 = Ca (NO3) 2 + C2H2.

C2H2 + 2H2O = Ca (OH) 2 + C2H2.

Ca (OH) 2 + 2HCl = CaCl2 + 2H2O.

n (Ca (OH) 2) = ½ n (HNO3) = 0.14: 2 = 0.07 mol.

n (Ca (OH) 2) = 0.2 – 0.07 = 0.13 mol.

For 1 mol of Ca (OH) 2, there are 2 mol of HCl.

The substances are in quantitative ratios of 1: 2.

The amount of HCl is 2 times more than the amount of Ca (OH) 2.

n (HCl) = 2 n (Ca (OH) 2) = 0.13 × 2 = 0.26 mol.

Find the mass of HCl.

М (HCl) = 36.5 g / mol.

m = n × M.

m = 36.5 g / mol × 0.26 mol = 9.49 g.

Find the mass of the HCl solution.

m (solution) = m (substance): W) × 100%.

m (solution) = (9.49 g: 20%) × 100% = 47.45 g.

Find the volume of HCl.

V = m: p.

V = 47.45 g: 1.10 g / ml = 43.14 ml.

Answer: V = 43.14 ml.