How many grams of silicic acid can be obtained from 50g. sodium silicate and 18.25 g hydrochloric acid?
| November 19, 2020 | education
Na2SiO3 + 2HCl = 2NaCl + H2SiO3 (sediment)
1.amount (Na2SiO3) = 50g / 122 g / mol = 0.4 mol – lacking
2.amount (HCl) = 18.25 g / 36.5 g / mol = 0.5 mol – in excess
The calculation is based on a disadvantage.
3.amount (Na2SiO3) = amount (H2SiO3) = 0.4 mol
4.m (H2SiO3) = 0.4 mol • 78 g / mol = 31.2 g.
Answer: m (H2SiO3) = 31.2 g.
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