Question

How many grams of NH3 are needed to react with 53.3g of K2PtCl4?

Answer 1

Answer on Question #47936, Chemistry, Other

**Task:**

How many grams of $\mathrm{NH}_3$ are needed to react with $53.3\ \mathrm{g}$ of $\mathrm{K}_2\mathrm{PtCl}_4$?

**Answer:**

$$\mathrm{K}_2\mathrm{PtCl}_4 + \mathrm{NH}_3 = 2\ \mathrm{KCl} + \mathrm{Pt}(\mathrm{NH}_3)\mathrm{Cl}_2$$ $$v = \frac{m}{M}$$

where m = mass, grams;

M = molar mass, gram/mol.

$$M(\mathrm{K}_2\mathrm{PtCl}_4) = 415\ \mathrm{g/mol} \quad M(\mathrm{NH}_3) = 17\ \mathrm{g/mol}$$ $$v(\mathrm{K}_2\mathrm{PtCl}_4) = \frac{53.3}{415} = 0.128\ \text{moles}$$ $$v(\mathrm{K}_2\mathrm{PtCl}_4) = v(\mathrm{NH}_3) = 0.128\ \text{moles}$$ $$m(\mathrm{NH}_3) = v(\mathrm{NH}_3) \cdot M(\mathrm{NH}_3)$$ $$m(\mathrm{NH}_3) = 0.128 \cdot 17 = 2.18\ \mathrm{g}$$

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