Question #19122

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Question #19122

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Answer 1 · 2015-01-29T18:27:02

$5.25 g$ $"5.25 g"$ of $NaF$ $"NaF"$ will be produced (or $5 g$ $"5 g"$ if you round to 1 sig fig - the number of sig figs 0.5 has).

So, you have your balanced chemical equation, which I won't write again here. Notice that you have an $8:2$ $"8:2"$ , or better said, a $4:1$ $"4:1"$ mole ratio between $HF$ $"HF"$ and $NaF$ $"NaF"$ .

This means that $4 moles$ $"4 moles"$ of $HF$ $"HF"$ will produce $1 mole$ $"1 mole"$ of $NaF$ $"NaF"$ . Since you were given the number of $HF$ $"HF"$ moles, and told that $N a_{2} S i O_{3}$ $Na_2SiO_3$ was not a limiting reagent, you can determine the number of $NaF$ $"NaF"$ moles to be

$0.5 moles HF \cdot \frac{1 mole NaF}{4 moles HF} = 0.125 moles$ $"0.5 moles HF" * ("1 mole NaF")/("4 moles HF") = "0.125 moles"$

$NaF$ $"NaF"$ has a molar mass of $42.0 g/mol$ $"42.0 g/mol"$ , which means that the mass produced will be

$0.125 moles \cdot \frac{42.0 g}{1 mole} = 5.25 g NaF$ $"0.125 moles" * ("42.0 g")/("1 mole") = "5.25 g NaF"$

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Question #19122

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