Answer:
just see it it will help trust me its my school work
Explanation:
Answer : 51.8 g of nitrogen are needed to produce 100 grams of ammonia gas.
Solution : Given,
Mass of
= 100 g
Molar mass of
= 27 g/mole
Molar mass of
= 28 g/mole
First we have to calculate moles of
.

The given balanced chemical reaction is,

From the given reaction, we conclude that
2 moles of
produced from 1 mole of 
3.7 moles of
produced from
of 
Now we have to calculate the mass of
.
Mass of
= Moles of
× Molar mass of 
Mass of
= 1.85 mole × 28 g/mole = 51.8 g
Therefore, 51.8 g of nitrogen are needed to produce 100 grams of ammonia gas.
Answer is: 2) 117g.
2Na + Cl₂ → 2NaCl
Step 1: calculate amount of substance of sodium and chlorine.
n(Na) = m(Na)÷M(Na) = 46g ÷ 23 g/mol = 2 mol.
n(Cl₂) = m(Cl₂)÷M(Cl₂) = 71g ÷ 71 g/mol = 1 mol.
Step 2: calculate amount of substance and mass of sodium-chloride.
Because both sodium and chlorine react completely, we can use both n to compare with n of NaCl.
n(Na) : n(NaCl) = 2:2, 2 mol : n(NaCl) = 2:2
n(NaCl) = 2mol, m(NaCl) = 2mol ·5805 g/mol = 117 g.
Answer:
25.6g de HF son producidos
Explanation:
<em>...¿Cuánto HF es producido?</em>
Para resolver este problema debemos convertir la masa de cada reactivo a moles usando su masa molar. Como la reacción es 1:1, el reactivo con menor número de moles es el reactivo limitante. Con las moles del reactivo limitante podemos obtener las moles de HF y su masa así:
<em>Moles CaF2:</em>
Masa molar:
1Ca = 40g/mol
2F = 19*2 = 38g/mol
40+38 = 78g/mol
50g CaF2 * (1mol/78g) = 0.641 moles CaF2
<em>Moles H2SO4:</em>
Masa molar:
2H = 2g/mol
1S = 32g/mol
4O = 64g/mol
98g/mol
100g H2SO4 * (1mol / 98g) = 1.02 moles H2SO4
Como las moles de CaF2 < Moles H2SO4: CaF2 es reactivo limitante.
<em>Moles HF usando la reacción:</em>
0.641 moles CaF2 * (2mol HF / 1mol CaF2) = 1.282 moles HF
<em>Masa HF:</em>
Masa molar:
1g/mol + 19g/mol = 20g/mol
1.282 moles HF * (20g/mol) =
<h3>25.6g de HF son producidos</h3>
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