How much naalo2 (sodium aluminate) is required to produce 5.01 kg of na3alf6?

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

Google it that might work

m=9

step-by-step explanation:

7m+8=71 subtract   8 from 71 which is 63 than divide it by 7m   equal 9.