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hope that's right and hope i helped
Answer:
0.15215407
Explanation:
Im not 100% sure if this is correct!
One should be tested at once. You never want to test more because you will mix up your data
Answer:
81.97 g of NaAl(OH)₄
Solution:
The reaction for the preparation of Sodium Aluminate from Aluminium Metal and NaOH is as follow,
2 NaOH + 2 Al + 6 H₂O → 2 NaAl(OH)₄ + 3 H₂
According to this equation,
2 Moles of NaOH produces = 163.94 g (2 mole) of NaAl(OH)₄
So,
1 Mole of NaOH will produce = X g of NaAl(OH)₄
Solving for X,
X = (1 mol × 163.94 g) ÷ 2 mol
X = 81.97 g of NaAl(OH)₄
