Question

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

Answer 1

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

Step-by-step 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]}=(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:
`(3)/(1)* (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* \text{molecular mass of}NaAlO_2}`

`=71.58 mol* 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]`.

Answer 2

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