How much base and acid do you need to take to obtain 200 grams of salt according to the formula NaOh + H3PO4.

Let’s implement the solution:
1. In accordance with the condition of the problem, we write the equation:
1 mol 1 mol;
3NaOH + H3PO4 = Na3PO4 + 3H2O – ion exchange reaction, sodium phosphate was formed;
2. Calculations by formulas:
M (NaOH) = 39.9 g / mol;
M (H3PO4) = 97.9 g / mol;
M (Na3PO4) = 117.9 g / mol.
3. Determine the number of moles of salt, if the mass is known:
M (Na3PO4) = m / M = 200 / 117.8 = 1.69 mol.
4. Proportions:
X mol (NaOH) – 1.69 mol (Na3PO4);
-3 mol -1 mol hence, X mol (NaOH) = 3 * 1.69 / 1 = 5 mol;
Y (H3PO4) = 1.69 mol since the amount of these substances according to the equation is 1 mol.
5. Find the mass of the starting materials:
m (NaOH) = Y * M = 5 * 39.9 = 199.5 g;
m (H3PO4) = Y * M = 1.69 * 97.9 = 165.45 g.
Answer: the mass of sodium hydroxide is 199.5 g, the mass of the acid is 165.45 g.