(1)0.5mol H2O的質(zhì)量為______,其中含有______個(gè)水分子,共有______個(gè)原子.
(2)質(zhì)量都是50g的 HCl、NH3、CO2、O2四種氣體中,含有分子數(shù)目最少的是______,在相同溫度和相同壓強(qiáng)條件下,體積最大的是______,體積最小的是______.
(3)配制50mL 0.2mol/L CuSO4溶液,需要CuSO4______g;需要CuSO4?5H2O______g.
(4)中和相同體積、相同物質(zhì)的量濃度的NaOH溶液,并使其生成正鹽,需要相同物質(zhì)的量濃度的HCl、H2SO4、H3PO4溶液的體積比為______.
解:(1)m(H
2O)=0.5mol×18g/mol=9g,N(H
2O)=0.5mol×N
A=3.01×10
23,N(H)+N(O)=3×N(H
2O)=9.03×10
23,
故答案為:9;3.01×10
23;9.03×10
23;
(2)由n=
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=
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可知:相同質(zhì)量時(shí),物質(zhì)的相對(duì)分子質(zhì)量越大,則分子數(shù)越小,CO
2的相對(duì)分子質(zhì)量最大,
則質(zhì)量都是50g的 HCl、NH
3、CO
2、O
2四種氣體中,含有分子數(shù)目最少的是CO
2,
n=
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=
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可知:相同質(zhì)量時(shí),物質(zhì)的相對(duì)分子質(zhì)量越大,則物質(zhì)的量越小,體積越小,則質(zhì)量都是50g的 HCl、NH
3、CO
2、O
2四種氣體中,體積最大的是NH
3,體積最小的是CO
2,
故答案為:CO
2;NH
3;CO
2;
(3)50mL 0.2mol/L CuSO
4溶液中:n(CuSO
4)=0.05L×0.2mol/L=0.01mol,
則:m(CuSO
4)=0.01mol×160g/mol=1.6g,
m(CuSO
4?5H
2O)=0.01mol×250g/mol=2.5g,
故答案為:1.6; 2.5;
(4)根據(jù)反應(yīng)H
++OH
-=H
2O可知,中和相同體積、相同物質(zhì)的量濃度的NaOH溶液,并使其生成正鹽,
則需:n(HCl):n(H
2SO
4):n(H
3PO
4)=1:
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:
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=6:3:2,
需要相同物質(zhì)的量濃度的HCl、H
2SO
4、H
3PO
4溶液的體積比為6:3:2,
故答案為:6:3:2.
分析:(1)根據(jù)n=
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=
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結(jié)合分子組成計(jì)算;
(2)n=
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=
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=
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計(jì)算;
(3)n=
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=cV計(jì)算;
(4)根據(jù)反應(yīng)H
++OH
-=H
2O,結(jié)合分子組成計(jì)算.
點(diǎn)評(píng):本題考查物質(zhì)的量的相關(guān)計(jì)算,題目難度中等,(4)為易錯(cuò)點(diǎn),注意三種酸的特點(diǎn).