Question

How much NaAlO2 (sodium aluminate) is required to produce 5.01 kg of Na3AlF6?

Answer 1

Balance Chemical equation is as follow,

                        3NaAlO₂ + 6HF   →   Na₃AlF₆ + 3H₂O + Al₂O₃

According to this equation,

      209.94 g of Na₃AlF₆ is produced by reacting = 245.91 g of NaAlO₂

So,

      5010 g of Na₃AlF₆ will be produced by reacting = X grams of NaAlO₂

Solving for X,
                        X  =  (5010 g × 245.91 g) ÷ 209.94 g

                        X  =  5868 g
Or,
                        X  =  5.868 Kg of NaAlO₂

Answer 2

Answer:5.86 kg sodium aluminate is required to produce 5.01 kg of $Na_3[AlF_6]$.

Explanation:

$3NaAlO_2+6HF\rightarrow Na_3[AlF_6]+Al_2O_3+3H_2O$

Mass of $Na_3[AlF_6]=5.01kg=5010g$(1kg=1000g)

$\text{Moles of }Na_3[AlF_6]=\frac{\text{Given mass of}Na_3[AlF_6]}{\text{molecular mass of}Na_3[AlF_6]}=\frac{5010 g}{ 209.94g/mol}=23.86 moles$

According to reaction, 1 mol of $Na_3AlF_6$ is obtained from 3 moles of $NaAlO_2$ , then 23.86 moles of $Na_3AlF_6$ will be obtained from: $\frac{3}{1}\times (23.86 moles)=71.58$ moles of $NaAlO_2$

Mass of sodium aluminate is required to produce 5.01 kg of $Na_3[AlF_6]$.

$\text{Moles of }NaAlO_2=\frac{\text{Mass of}NaAlO_2}{\text{molecular mass of}NaAlO_2}=\frac{\text{Mass of}NaAlO_2}{81.97 g/mol}$

$\text{Mass of}NaAlO_2}=\text{Moles of }NaAlO_2\times \text{molecular mass of}NaAlO_2}$

$=71.58 mol\times 81.97 g/mol=5867.41g=5.86741 kg\approx5.86 kg$

5.86 kg sodium aluminate is required to produce 5.01 kg of $Na_3[AlF_6]$.