na2sio3 + 8 hf --> h2sif6 + 2 naf + 3 h2o according to the balanced chemical equation above, how many grams…

na2sio3 + 8 hf --> h2sif6 + 2 naf + 3 h2o according to the balanced chemical equation above, how many grams of naf form when 0.486 mol of hf reacts with excess na2sio3?

na2sio3 + 8 hf --> h2sif6 + 2 naf + 3 h2o according to the balanced chemical equation above, how many grams of naf form when 0.486 mol of hf reacts with excess na2sio3?

Answer

Explanation:

Step1: Determine mole - ratio

From the balanced equation $Na_2SiO_3 + 8HF\rightarrow H_2SiF_6+2NaF + 3H_2O$, the mole - ratio of $HF$ to $NaF$ is $8:2$ or $4:1$.

Step2: Calculate moles of $NaF$

If $n_{HF}=0.486$ mol, and $\frac{n_{NaF}}{n_{HF}}=\frac{2}{8}=\frac{1}{4}$, then $n_{NaF}=\frac{0.486}{4}=0.1215$ mol.

Step3: Calculate mass of $NaF$

The molar mass of $NaF$ is $M_{NaF}=22.99 + 18.998=41.988$ g/mol. Using the formula $m = n\times M$, we have $m_{NaF}=0.1215\times41.988\approx5.10$ g.

Answer:

$5.10$ g