Answer:
23.8g of sodium phosphate are formed
Explanation:
Based on the reaction of sodium, Na, with phosphoric acid, H₃PO₄:
3Na + H₃PO₄ → Na₃PO₄ + 3/2 H₂
<em>3 moles of sodium produce 1 mole of sodium phosphate</em>
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To solve this question we must find the moles of sodium in 10g. Using the chemical reaction we can find the moles -And the mass- of sodium phosphate produced, as follows:
<em>Moles Na -Molar mass: 22.99g/mol-</em>
10g * (1mol / 22.99g) = 0.435 moles Na
<em>Moles Na₃PO₄:</em>
0.435 moles Na * (1mol Na₃PO₄ / 3mol Na) = 0.145 moles Na₃PO₄
<em>Mass Na₃PO₄ -Molar mass: 163.94g/mol-</em>
0.145 moles Na₃PO₄ * (163.94g/mol) =
<h3>23.8g of sodium phosphate are formed</h3>
To get the number of liters of water vapor produced from the combustion of methane gas, we just need the stoichiometric ratio of water to methane which is 2:1. So the number of liters of water vapor from 13.3 liters of methane is 26.6 liters.
Answer:
cant read the questions...
Explanation:
13.82 g / 180.16 g/mol = .07671 moles
.07671 moles / (86.18 g / 1000 g/kg) = .8901 molal
Let me know if you have any further questions!
The mass of GeCl4 that would be formed will be 1.51 g
<h3>Stoichiometric calculations</h3>
From the balanced equation of the reaction:
Ge (s) + 2Cl2 (g) →GeCl4
The mole ratio of Ge and Cl2 is 1:2.
Mole of 1.0 g Ge: 1/72.64
= 0.01377 mole
Mole of 1.0 g Cl2 = 1/71
= 0.01408 mole
Thus, Cl2 is limiting.
Mole ratio of Cl2 and GeCl4 = 2:1
Equivalent mole of GeCl4 = 0.01408/2
= 0.0070 mole
Mass of 0.0070 mole GeCl4 = 0.0070 x 214.4
= 1.51 g
More on stoichiometric calculations can be found here: brainly.com/question/8062886
