1)For the reaction, calculate how many grams of the product form when 16.4 g of O2 completely reacts.
Assume that there is more than enough of the other reactant.
4Cr(s)+3O2(g)---->2Cr2O3(s)
2)For the reaction, calculate how many grams of the product form when 16.4 g of Sr completely reacts.
Assume that there is more than enough of the other reactant.
2Sr(s)+O2(g)---->2SrO(s)
3 Answers
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1) molar mass of Cr2O3=2*52.00+3*16.00=152g
16.4gofO2*1molO/16.00gO * 2molCr2O3/3molO2 * 152gCr2O3/1molCr2O3=104gCr2O3
2)molar mass of SrO=87.62+16.00=103.62g
16.4gSr * 1molSr/87.62Sr * 2molSrO/2molSr * 103.62gSrO/1molSrO = 19.4 g SrO
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What you are assuming is that reaction goes to completion and that all of the the reactant in question is used up. You need to do some unit analysis and find out the moles of reactant used, then set up a mol to mol ration with the product then find the amount used via molecular weight.
1) 16.4g 02 * 1 mol O2/15.99g * 2 mol Cr2O3/ 1 mol 02 * 151.92g Cr2O3/ 1mol Cr2O3 = 311.63g.
2) 16.4g Sr * 1mol Sr/ 87.62g * 2 mol SrO/ 2 mol Sr * 103.61g / 1 mol SrO = 19.39g.
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#1 Ans
A. See what relevant information the balanced equation tells you.
It tells you 3O2 forms 2Cr2O3
B. Use moles
3 moles oxygen forms 2 moles Cr2O3
C. Work out mass of relevant moles
O2 = 16 x 2 = 32g. Cr2O3 = (52 x 2) + 48 = 152g
D. Substitute masses in B
96g O2. Forms. 304g Cr2O3
E. Use simple proportion
16.4g O2 form 304/96 x 16.4g Cr2O3 answer 51.9grams with appropriate significant figures