1) Find the number of mols of HCl in 5.2 liters of 4.0M solution:
n = M*V(L) = 4.0 mol/L * 5.2 L = 20.8 mol
2) Find the number of mols of Mg that will react with 20.8 mol of HCl, using the coefficients of the balanced equation
[1mol Mg / 2 mol HCl] * 20.8 mol HCl = 10.4 mol Mg
3) Transform mol to mass using the atomic mass:
10.4 mol Mg * 24.3 g/mol = 252.7 g of Mg.
I think the answer is ‘repulsion’
The answer is 615.91 grams of <span>n2f4
Solution:
225g F2 x [(1molF2)/(38gramsF2)] x [</span>(1molF2)/(1molN2F4)] x [(104.02 grams N2F4)/(1molN2F4)]
=615.91 grams
91 grams of sodium azide required to decompose and produce 2.104 moles of nitrogen.
Explanation:
2NaN3======2Na+3N2
This is the balanced equation for the decomposition and production of sodium azide required to produce nitrogen.
From the equation:
2 moles of NaNO3 will undergo decomposition to produce 3 moles of nitrogen.
In the question moles of nitrogen produced is given as 2.104 moles
so,
From the stoichiometry,
3N2/2NaN3=2.104/x
= 3/2=2.104/x
3x= 2*2.104
= 1.4 moles
So, 1.4 moles of sodium azide will be required to decompose to produce 2.104 moles of nitrogen.
From the formula
no of moles=mass/atomic mass
mass=no of moles*atomic mass
1.4*65
= 91 grams of sodium azide required to decompose and produce 2.104 moles of nitrogen.
