Avogadro's law and molar volume of gases

Apply Avogadro's law that equal gas volumes at the same temperature and pressure contain equal numbers of molecules, use the molar volume of 22.7 dm³ mol⁻¹ at STP, and use mole ratios from equations to calculate reacting gas volumes.

Zadania sprawdzające tę umiejętność: 7 Otwórz w wyszukiwarce z filtrami
Zadanie 24Paper 1A, maj 2026
wzorowane1 pktzamkniętełatwe (szac.)

Which volume of oxygen is required to react fully with 40.0 cm3 of butane to ensure complete combustion? 2CX4HX10(g)+13OX2(g)8COX2(g)+10HX2O(g)\ce{2C4H10(g) + 13O2(g) -> 8CO2(g) + 10H2O(g)} A. 520 cm3 B. 260 cm3 C. 130 cm3 D. 40.0 cm3

Zadanie 3Paper 1A, maj 2025, TZ3
wzorowane1 pktzamkniętełatwe (szac.)

What volume of propane gas, in cm3\mathrm{cm^3}, must be burned completely in oxygen to give 90 cm3\mathrm{cm^3} of carbon dioxide, all volumes being measured at the same temperature and pressure? CX3HX8(g)+5OX2(g)3COX2(g)+4HX2O(g)\ce{C3H8(g) + 5O2(g) -> 3CO2(g) + 4H2O(g)} A. 18 B. 30 C. 90 D.…

Zadanie 27Paper 1A, maj 2025, TZ1
wzorowane1 pktzamknięteśrednie (szac.)

30 cm3\mathrm{cm^3} of butane was completely burned in 250 cm3\mathrm{cm^3} of oxygen according to the following equation. 2CX4HX10(g)+13OX2(g)8COX2(g)+10HX2O(l)\ce{2C4H10(g) + 13O2(g) -> 8CO2(g) + 10H2O(l)} What volume of oxygen remains unreacted, measured at the original conditions? A. 55…

Zadanie 25Paper 1A, specimen 2025
wzorowane1 pktzamknięteśrednie (szac.)

10.0 cm310.0\ \mathrm{cm^3} of a gaseous hydrocarbon, CXxHXy\ce{C_xH_y}, is burned completely in exactly the required volume of oxygen. The gaseous products occupy 70.0 cm370.0\ \mathrm{cm^3}. This falls to 30.0 cm330.0\ \mathrm{cm^3} once the water vapour present condenses, with…

Zobacz wszystkie 7 w wyszukiwarce z filtrami