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Answer:
3.52g
Explanation:
Zr = 90 g/mol
Si = 28 g/mol
O = 16 g/mol
ZrSiO4 = 90 + 28 + (16* 4) = 182g/mol
Mass = Numberof moles * Molar mass
Mass = 1mol * 182g/mol = 182g
In one mole of ZrSiO4, there are 4 oxygen atoms, hence the mass is given as;
4 * 16 = 64g
Hence, 64g of oxygen is present in 182g of ZrSiO4.
10g would contain x
64 = 182
x = 10
x = ( 10 * 64 ) / 182
x = 640 / 182 = 3.52g
At the first reaction when 2HBr(g) ⇄ H2(g) + Br2(g)
So Kc = [H2] [Br2] / [HBr]^2
7.04X10^-2 = [H2][Br] / [HBr]^2
at the second reaction when 1/2 H2(g) + 1/2 Br2 (g) ⇄ HBr
Its Kc value will = [HBr] / [H2]^1/2*[Br2]^1/2
we will make the first formula of Kc upside down:
1/7.04X10^-2 = [HBr]^2/[H2][Br2]
and by taking the square root:
∴ √(1/7.04X10^-2)= [HBr] / [H2]^1/2*[Br]^1/2
∴ Kc for the second reaction = √(1/7.04X10^-2) = 3.769