How many grams of NaF form when .508 mol of HF react with excess Na2SiO3?

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04-09-2017
Chemistry

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How many grams of NaF form when .508 mol of HF react with excess Na2SiO3?

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Аноним Аноним · Answer 1 · 2017-09-04T19:01:48+02:00

Аноним Аноним

04-09-2017

2HF + Na2SiO3 ----> 2NaF + H2SiO3 is the balanced equation.

2 moles of HF form 2 moles of NaF

therefore 0.508 Mol HF forms 0.508 mol of NaF

In grams this is 0.508( rel atomic mass of Na + RAM of F)
= 0.508 (22.99+19)
= 21.33 grams

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