Question

sodium nitrate heated in the presence of an excecesd of H2 from H20 according to step process
2NaNO3-2NaNO2+O2
2H+O2-2H2O
how many grams of NaNO3 are required to from 9g of H2O?​

Answer 1

1 kgillion kilogram is required

Answer 2

42.5 grams of NaNO3 are required to produce 9g of H2O with the given two balanced chemical reactions.

Explanation:

• No. of moles of H20 in 9g = given weight/molecular weight = 9/18 = 0.5 moles. According to the balanced equation, 2 H2 + O2 -> 2 H2O, one mole of O2 can give 2 moles of H2O as the product. So for 0.5 moles of H2O, we need 0.25 moles of O2.

                                      2 moles H20 <- 1 mole O2

                                      1 mole H2O <- 0.5 mole O2

                                      0.5 mole H2O <- 0.25 mole O2        

• 2 NaNO3 -> 2 NaNO2 + O2 is balanced. We know 2 moles of NaNO3 are necessary to produce 1 mole of O2. So it requires 0.5 moles of NaNO3 to produce 0.25 moles of O2.    

                                     2 moles NaNO3 -> 1 mole O2

                                     1 mole NaNO3 -> 0.5 moles O2

                                      0.5 moles NaNO3 -> 0.25 moles O2

• No. of moles of NaNO3 =     $\frac{Weight of NaNO3(g)}{Molecular weight of NaNO3}$

           Weight of NaNO3(g) = No. of moles x Molecular weight

                  Weight of NaNO3(g) = 0.5 x 85

                                                  [Molecular weight of NaNO3= 23+14+3x16=85]

                  Weight of NaNO3(g) = 42.5 g