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Quantitative Chemistry: Using Moles to Balance Equations (DONE)

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卡片总数: 3内容版本: v4公开卡包更新时间: 8/1/2026

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#1
正面 (问题)

Balance the equation using moles 2g of hydrogen reacts with 71g of chlorine to make 73g hydrogen chloride. Ar H =1 Ar Cl = 35.5 H↓2 + Cl↓2 → HCl

背面 (解答)

Number of moles = mass (g) / relative formula mass H Ar = 1 x 2 = 2 number of moles 1 = 2g / 2 Cl Ar = 2 x 35.5 = 71 1 = 71g / 71 HCL = 35.5 + 1 = 36.5 73g / 36.5 = 2 1H↓2 + 1Cl↓2 → 2HCl we’ve now got to make certain that these bigger numbers are in the lowest possible ratio and to do this we divide all of them by the smallest number. So in this case the lowest number is one. 1 / 1 = 1, 1 / 1 = 1, 2 / 1 = 2 H↓2 + 1Cl↓2 → 2HCl (We don’t write ones in chemical equations)

#2
正面 (问题)

54g of aluminium reacts with 216g of iron oxide, forming 102g of aluminium oxide and 168g of iron. Balance the equation using moles. Ar Al = 27 Ar Fe = 56 Ar O = 16 Al + FeO → Al↓2O↓3 + Fe

背面 (解答)

2Al + 3FeO → Al↓2O↓3 + 3Fe

#3
正面 (问题)

1248g of barium chloride reacts with 684g of aluminium sulphate, forming 1398 of barium sulphate and 534g of aluminium chloride chloride. Balance the equation. Ba = 137. Cl = 35.5. S = 32. O = 16. Al = 27 BaCl↓2 + Al↓2(SO↓4)↓3 → BaSO↓4 + AlCl↓3

背面 (解答)

3BaCl↓2 + Al↓2(SO↓4)↓3 → 3BaSO↓4 + 2AlCl↓3