23.458 g
<h3>Further explanation</h3>
Given
Reaction
H2SO4 + 2NaNO2 → 2HNO2 + Na2SO4
24.14714 g of Na2SO4
Required
mass of NaNO2
Solution
mol Na2SO4 :
= 24.14714 : 142,04 g/mol
= 0.17
From the equation, mol NaNO2 :
= 2/1 x mol Na2SO4
= 2/1 x 0.17
= 0.34
Mass NaNO2 :
= 0.34 x 68,9953 g/mol
= 23.458 g
The sand provides a rough surface on top of the ice for the cars' tires to grip onto. It provides more friction. Salt melts the ice and often provides more friction (the disadvantage is that it eats concrete!).
Answer:
0.190L of hydrogen may be produced by the reaction.
Explanation:
Our reaction is:
3Mg + 2H₃PO₄ → Mg₃(PO₄)₂ + 3H₂
We need to determine the limting reactant. Let's find out the moles of each:
5.159×10²¹ atoms . 1 mol / 6.02×10²³ atoms = 0.00857 moles of Mg
55.23 g . 1 mol / 97.97 g = 0.563 moles of acid
2 moles of acid react to 3 moles of Mg
0.563 moles of acid may react to: (0.563 . 3) /2 = 0.8445 moles of Mg
Definetely the limting reactant is Mg.
As ratio is 3:3, 3 moles of Mg can produce 3 moles of hydrogen
Then, 0.00857 moles of Mg must produce 0.00857 moles of H₂
At STP, 1 mol of any gas occupies 22.4L
0.00857 mol . 22.4L / 1mol = 0.190L
I think it’s D. I’m not really shore....