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morpeh [17]
3 years ago
9

1 Srcl (aq) + 1 H,60,(aq) → 2 HCl(aq) + 1 Srso (s)

Chemistry
1 answer:
Mrac [35]3 years ago
3 0

Answer:

Mass = 245.72 g

Explanation:

Given data:

Mass of SrCl₂ react = ?

Mass of H₂SO₄ = 300.0 g

Solution:

SrCl₂  +  2H₂SO₄     →     2HCl + Sr(HSO₄)₂

Number of moles of H₂SO₄:

Number of moles = mass/molar mass

Number of moles =  300.0 g/ 98.079 g/mol

Number of moles = 3.1 mol

Now we will compare the moles of SrCl₂ and  H₂SO₄.

                   H₂SO₄         :       SrCl₂

                        2             :           1

                      3.1             :         1/2×3.1 = 1.55 mol

Mass of SrCl₂:

Mass = number of moles × molar mass

Mass = 1.55 mol × 158.53 g/mol

Mass = 245.72 g

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3 years ago
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4 years ago
Cuando se trata Ca3P2 con agua, los productos son Ca(OH)2 y PH3. Calcular el peso máximo obtenido al reaccionar 2 g de Ca3P2 con
anyanavicka [17]

Answer:

0.629 gramos de PH3 es la máxima de cantidad que puede ser producida.

Explanation:

¡Hola!

En este caso, dado que tenemos la siguiente reacción química, la cual se puede balancear directamente:

Ca_3P_2+6H_2O\rightarrow 3Ca(OH)_2+2PH_3

Podemos calcular la masa máxima de cualquier producto, digamos PH3, al comparar la masa de este, que 2 g the Ca3P2 y 1 g de H2O producen por separado y de acuerdo a la estequiometría:

2gCa_3P_2 *\frac{1molCa_3P_2 }{182.18gCa_3P_2 }*\frac{2molPH_3}{1molCa_3P_2 }  *\frac{34gPH_3}{1molPH_3 } =0.747gPH_3\\\\1gH_2O *\frac{1molH_2O }{18.02gH_2O}*\frac{2molPH_3}{6molH_2O}  *\frac{34gPH_3}{1molPH_3 } =0.629gPH_3

De este modo, infermos que solamente 0.629 gramos the PH3 pueden ser obtenidos al ser el agua el reactivo límite.

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3 years ago
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