dvhtlbz的部落格

標題:

Calculate the% by mass of calcium carbonate in the sample

發問:

• 點計subnet mask@1@
• 可以介紹皮膚專科醫生比我嗎
• $4000 出機一問@1@
• What is 午餐321-----
• 我唔識數學!!
• 請問密宗藏傳佛教喇嘛教與佛教明顯不同卻被忽視之處@1@
• 最近銅鑼灣sogo的郵筒-
• 求216和306的H.C.F@1@
• 砌機問題-
• PSP - Loco Roco

此文章來自奇摩知識+如有不便請留言告知

step1 11g of sample were placed in a beaker step 2120cm^3 of 2M nitric acid were added to react with the samplestep 3the excess acid was titrated against 1.60M sodium hydroxide solutionwith phenolphthalein as indicator. 17.5cm^3 of the alkali were required to reach the... 顯示更多 step1 11g of sample were placed in a beaker step 2 120cm^3 of 2M nitric acid were added to react with the sample step 3 the excess acid was titrated against 1.60M sodium hydroxide solution with phenolphthalein as indicator. 17.5cm^3 of the alkali were required to reach the endpoint. R.A.M:Ca=40.1,C=12.0,O=16.0

最佳解答:

Consider the titration of excess HNO3 (unreacted with the sample) against NaOH. HNO3 + NaOH → NaNO3- + H2O Mole ratio HNO3 : NaOH = 1 : 1 No. of moles of NaOH used = 1.6 x (17.5/1000) = 0.028 mol No. of moles of HNO3 unreacted with the sample = 0.028 mol Consider the reaction of HNO3 with the sample of impure CaCO3: (Assume the impurities do not cause any reaction.) CaCO3 + 2HNO3 → Ca(NO3)2 + H2O + CO2 Mole ratio CaCO3 : HNO3 = 1 : 2 Total number of HNO3 added = 2 x (120/1000) = 0.24 mol No. of moles of HNO3 unreacted = 0.028 mol No. of moles of HNO3 reacted = 0.24 - 0.028 = 0.212 mol No. of moles of CaCO3 used = 0.212 x (1/2) = 0.106 mol Molar mass of CaCO3 = 40 + 12 + 16x3 = 100 g mol-1 Mass of CaCO3 used = 0.106 x 100 = 1.06 g % by mass of CaCO3 in the sample = (1.06/1.1) x 100% = 96.36% =

其他解答:FAD2A23AB937987B