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hannaharsena
20.04.2020 •
Chemistry
5.943x10^24 molecules of H3PO4 will need how many grams of Mg(OH)2 in the reaction below? 3 Mg(OH)2 + 2 H3PO4 > 1 Mg3(PO4)2 + 6 H2O
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Ответ:
5.943 × 10²⁴ molecules of H₃PO₄ will react with 789.29 grams of Mg(OH)₂
Explanation:
Here we have
5.943 × 10²⁴ molecules of H₃PO₄ in a reaction with Mg(OH)₂
as follows
3Mg(OH)₂ + 2H₃PO₄ → Mg₃(PO₄)₂ + 6H₂O
Therefore 3 moles of Mg(OH)₂ react with 2 moles of H₃PO₄ to form 1 mole of Mg₃(PO₄)₂ and 6 moles of H₂O
5.943 × 10²⁴ molecules of H₃PO₄ which is equivalent to![\frac{5.943 \times 10^{24}}{6.02 \times 10^{23}} moles = 9.869 \, moles \, of \, H_3PO_4](/tpl/images/0611/6062/b3c1e.png)
Will react with 3/2×9.869 moles or 14.8 moles of Mg(OH)₂
One mole of Mg(OH)₂ weighs 58.3197 g/mol
Therefore, 5.943 × 10²⁴ molecules of H₃PO₄ will react with 14.8×58.3197 g or 789.29 grams of Mg(OH)₂.
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