A sample of methane (CH4) has a volume of 25 mL at a pressure of 0.80 atm. What is the volume of the gas at each of the followin
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The mass of nitric acid (HNO₃) required for the production of 2.8 g of nitroglycerin? (C₃H₅N₃O₉) is 2.33 g
We'll begin by calculating the mass of HNO₃ that reacted and the mass of C₃H₅N₃O₉ produced from the balanced equation. This can be obtained as follow:
C₃H₈O₃ + 3HNO₃ —> C₃H₅N₃O₉ + 3H₂O
Molar mass of HNO₃ = 1 + 14 + (16×3)
= 1 + 14 + 48
= 63 g/mol
Mass of HNO₃ from the balanced equation = 3 × 63 = 189 g
Molar mass of C₃H₅N₃O₉ = (12×3) + (1×5) + (14×3) + (16 × 9)
= 36 + 5 + 42 + 144
= 227 g/mol
Mass of C₃H₅N₃O₉ from the balanced equation = 1 × 227 = 227 g
<h3>SUMMARY:</h3>From the balanced equation above,
227 g of C₃H₅N₃O₉ were produced by 189 g of HNO₃.
Finally, we shall determine the mass of HNO₃ that will produce 2.8 g of C₃H₅N₃O₉. This can be obtained as follow:
From the balanced equation above,
227 g of C₃H₅N₃O₉ were produced by 189 g of HNO₃.
Therefore,
2.8 g of C₃H₅N₃O₉ will be produce by = = 2.33 g of HNO₃
Thus, the mass of nitric acid (HNO₃) required for the production of 2.8 g of nitroglycerin? (C₃H₅N₃O₉) is 2.33 g
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