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1 year ago

How many dm3 of oxygen at STP would be required to react completely with 38.8g of propane? (3 marks)

Chemistry

1 answer:

11Alexandr11 [23.1K]1 year ago

5 0

$98.56 dm^3$ of oxygen at STP would be required to react completely with 38.8g of propane.

Given that :

molar mass of propane = 44 g/mol

mass of propane = 38.8 g

∴ Moles present in 38.8 g of propane = $\frac{38.8}{44}$ = 0.88 mole

applying rule of balanced equations

1 mole of propane = 5 moles of oxygen

0.88 mole of propane = 5 * 0.88 = 4.4 moles of oxygen

Note : volume of 1 mole of oxygen at STP = $22.4 dm^3$

∴Total volume of oxygen required at STP = 22.4 * 4.4 = $98.56 dm^3$

Hence we can conclude that the volume of oxygen at STP required to react completely $98.56 dm^3$

Learn more : brainly.com/question/16998374

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2 years ago

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Read 2 more answers

How many moles of MgO are produced when .250 mol of Mg reacts completely with O2

nignag [31]

Answer:

0.250 moles of MgO are produced when 0.250 mol of Mg reacts completely with O₂

Explanation:

In first place, the balanced reaction between Mg and O₂ is:

2 Mg + O₂ ⇒ 2 MgO

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of reactants and products participate in the reaction:

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Then you can apply the following rule of three: if by reaction stoichiometry 2 moles of Mg produce 2 moles of MgO, 0.250 moles of Mg, how many moles of MgO will they form?

$moles of MgO=\frac{0.250 moles of Mg*2 moles of MgO}{2 moles of Mg}$

moles of MgO= 0.250

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