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calculated?","back":"By summing the relative atomic masses of all atoms in the formula.","mcqOptions":null},{"id":"a62352a4-e753-4aa5-b07a-4f2bc6ecb184","cardType":"basic","points":1,"front":"What equation converts mass into moles?","back":"$$n = \\dfrac{m}{M_r}$","mcqOptions":null},{"id":"08a739ce-bea6-45af-84e9-3dbfb08ae028","cardType":"basic","points":1,"front":"What equation converts moles into mass?","back":"$$m = n \\times M_r$","mcqOptions":null},{"id":"3ea60c2b-47b5-4190-a13a-e05bfdb1fc93","cardType":"basic","points":2,"front":"Why must mass be converted into moles before using an equation ratio?","back":"Balancing numbers apply to moles, not directly to masses.","mcqOptions":null},{"id":"50903894-1694-442a-a916-6f551bb0b5a9","cardType":"basic","points":1,"front":"What is the first step when finding one mass from another?","back":"Convert the given mass into moles using $n = \\dfrac{m}{M_r}$.","mcqOptions":null},{"id":"28811a11-b117-4507-8d4f-d3e2da4fb86d","cardType":"basic","points":1,"front":"What is the second step when finding one mass from another?","back":"Use the balanced equation's mole ratio to find the required moles.","mcqOptions":null},{"id":"d41c8feb-01b2-442b-830a-df5e84e85584","cardType":"basic","points":1,"front":"What is the final step when finding one mass from another?","back":"Convert the required moles into mass using $m = n \\times M_r$.","mcqOptions":null},{"id":"9cead2a5-9838-403b-988d-a7f4129b77b1","cardType":"basic","points":2,"front":"In $2\\text{Mg} + \\text{O}_2 \\rightarrow 2\\text{MgO}$, how many moles of oxygen react with $0.25\\ \\text{mol}$ of magnesium?","back":"$0.125\\ \\text{mol}$ of oxygen.","mcqOptions":null},{"id":"0f426d40-1430-482a-bc3f-0187facec84f","cardType":"basic","points":2,"front":"In $2\\text{Mg} + \\text{O}_2 \\rightarrow 2\\text{MgO}$, how many moles of magnesium oxide form from $0.25\\ \\text{mol}$ of magnesium?","back":"$0.25\\ \\text{mol}$, because the Mg:MgO ratio is $2:2$.","mcqOptions":[{"text":"$0.25\\ \\text{mol}$, because the Mg:MgO ratio is $2:2$.","correct":true,"explanation":"The balanced equation has equal coefficients for Mg and MgO, giving a 1:1 mole ratio. Therefore 0.25 mol of magnesium forms 0.25 mol of magnesium oxide."},{"text":"$0.125\\ \\text{mol}$, because the Mg:MgO ratio is $2:1$.","correct":false,"explanation":"The 2:1 ratio applies to magnesium and oxygen, not magnesium and magnesium oxide. Mg and MgO have a 2:2 ratio, so the product amount is 0.25 mol."},{"text":"$0.50\\ \\text{mol}$, because the Mg:MgO ratio is $1:2$.","correct":false,"explanation":"This reverses the product relationship. The coefficients for Mg and MgO are both 2, so 0.25 mol of Mg produces 0.25 mol of MgO."},{"text":"$0.75\\ \\text{mol}$, because the Mg:MgO ratio is $2:3$.","correct":false,"explanation":"The equation contains coefficient 2 for both magnesium and magnesium oxide, not 3 for MgO. Their 1:1 ratio keeps the amount at 0.25 mol."}]},{"id":"0e18c03c-fc7f-41ea-9a6b-4758a83b3ebc","cardType":"basic","points":3,"front":"What mass of magnesium oxide forms when $6\\ \\text{g}$ of magnesium burns? Use $M_r(\\text{Mg}) = 24$ and $M_r(\\text{MgO}) = 40$.","back":"$10\\ \\text{g}$ MgO: $6/24 = 0.25\\ \\text{mol}$, then $0.25 \\times 40 = 10\\ \\text{g}$.","mcqOptions":null},{"id":"2bfde2ff-32b1-4df5-9cce-aa629de9a70a","cardType":"basic","points":3,"front":"What mass of water forms from $4\\ \\text{g}$ of hydrogen in $2\\text{H}_2 + \\text{O}_2 \\rightarrow 2\\text{H}_2\\text{O}$? Use $M_r(\\text{H}_2) = 2$ and $M_r(\\text{H}_2\\text{O}) = 18$.","back":"$36\\ \\text{g}$ water: $4/2 = 2\\ \\text{mol H}_2$, giving $2\\ \\text{mol H}_2\\text{O}$, then $2 \\times 18 = 36\\ \\text{g}$.","mcqOptions":null},{"id":"680e1bea-1e73-454c-98e5-c4126211e806","cardType":"cloze","points":null,"front":"The large numbers before formulae in a balanced equation give the [ ].","back":"The large numbers before formulae in a balanced equation give the mole ratio.","mcqOptions":null},{"id":"0dc0c007-546c-43a2-8643-c90f5af69302","cardType":"cloze","points":null,"front":"The equation for converting mass into moles is $\\text{[...]}$.","back":"The equation for converting mass into moles is $\\boxed{n = \\dfrac{m}{M_r}}$.","mcqOptions":null},{"id":"9ddc0d48-f469-4ee3-8765-701479782d8c","cardType":"cloze","points":null,"front":"The equation for converting moles into mass is $\\text{[...]}$.","back":"The equation for converting moles into mass is $\\boxed{m = n \\times M_r}$.","mcqOptions":null},{"id":"f9c95798-2d64-45bd-be88-42f97d3d3c4c","cardType":"cloze","points":null,"front":"In $2\\text{H}_2 + \\text{O}_2 \\rightarrow 2\\text{H}_2\\text{O}$, two moles of hydrogen react with [ ].","back":"In $2\\text{H}_2 + \\text{O}_2 \\rightarrow 2\\text{H}_2\\text{O}$, two moles of hydrogen react with one mole of oxygen.","mcqOptions":null},{"id":"226435aa-8dfe-4f77-b650-021478a26135","cardType":"cloze","points":null,"front":"In $2\\text{Mg} + \\text{O}_2 \\rightarrow 2\\text{MgO}$, the mole ratio of magnesium to magnesium oxide is $\\text{[...]}$.","back":"In $2\\text{Mg} + \\text{O}_2 \\rightarrow 2\\text{MgO}$, the mole ratio of magnesium to magnesium oxide is $\\boxed{2:2}$.","mcqOptions":null}]}],"examCountdown":{"examDate":"2027-05-17","paperLabel":"Paper 1"},"heading":"3.2.2 Amounts of substances in equations (HT only)","title":"3.2.2 Amounts of substances in equations (HT only) Flashcards","topbar":{"backHref":"/gcse/chemistry/aqa/flashcards"},"topicSidebar":"$L37"}]}],"topicId":"94aa471e-beec-5203-8376-9e173b216da6","walled":"$L38"}],"$L39","$L3a"]