How many particle or molecules of $N a_{2} C O_{3}$ $Na_2CO_3$ are produced when 10g of $N a O H$ $NaOH$ is used? $22.3 x 10^{24}$ $22.3 x 10^24$ particles of $N a O H$ $NaOH$ are needed to produce how many liters of $H_{2} O$ $H_2O$ ? How many particles/molecules of $N a O H$ $NaOH$ are from 30 gram of $N a O H$ $NaOH$ ?

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How many particle or molecules of $N a_{2} C O_{3}$ $Na_2CO_3$ are produced when 10g of $N a O H$ $NaOH$ is used? $22.3 x 10^{24}$ $22.3 x 10^24$ particles of $N a O H$ $NaOH$ are needed to produce how many liters of $H_{2} O$ $H_2O$ ? How many particles/molecules of $N a O H$ $NaOH$ are from 30 gram of $N a O H$ $NaOH$ ?

$2 N a O H (s) + C O_{2} (g) \to N a_{2} C O_{3} (s) + H_{2} O (l)$ $2NaOH(s) + CO_2(g) -> Na_2CO_3(s) + H_2O(l)$

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Answer 1 · 2017-08-02T13:48:30

a) $7.5 \times 10^{22}$ $7.5 xx 10^22$ molecules of $N a_{2} C O_{3}$ $Na_2CO_3$ are produced
b)cannot be answered as stoichiometry deals with gases not liquids and water is liquid
c)There are $4.5 \times 10^{23}$ $4.5xx 10^23$ molecules of $N a O H$ $NaOH$ in $30 g$ $30 g$

Explanation:

$2 N a O H (s) + C O_{2} (g) \to N a_{2} C O_{3} (s) + H_{2} O (l)$ $2NaOH(s) + CO_2(g) -> Na_2CO_3(s) + H_2O(l)$

Molar mass of $N a O H = 40$ $NaOH = 40$

$2 \times 40 g$ $2xx40 g$ of $N a O H$ $NaOH$ produces $6.02 \times 10^{23}$ $6.02 xx 10^23$ molecules of $N a_{2} C O_{3}$ $Na_2CO_3$
$:.$ $10 g$ of $NaOH$ produces $(6.02xx10^23xx10)/(2xx40)$ molecules of $Na_2CO_3$
$= 7.5 xx 10^22$ molecules of $Na_2CO_3$

$40g$ of $NaOH$ contains $6xx10^23$ molecules
$:.$ $30g$ of $NaOH$ contains $(30xx6xx10^23)/40$ molecules
$=4.5xx10^23$ molecules

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$2 N a O H (s) + C O_{2} (g) \to N a_{2} C O_{3} (s) + H_{2} O (l)$ $2NaOH(s) + CO_2(g) -> Na_2CO_3(s) + H_2O(l)$

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How many particle or molecules of Na_2CO_3Na2CO3Na_2CO_3 are produced when 10g of NaOHNaOHNaOH is used? 22.3 x 10^2422.3x102422.3 x 10^24 particles of NaOHNaOHNaOH are needed to produce how many liters of H_2OH2OH_2O? How many particles/molecules of NaOHNaOHNaOH are from 30 gram of NaOHNaOHNaOH?

2NaOH(s) + CO_2(g) -> Na_2CO_3(s) + H_2O(l)2NaOH(s)+CO2(g)→Na2CO3(s)+H2O(l)2NaOH(s) + CO_2(g) -> Na_2CO_3(s) + H_2O(l)

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$2 N a O H (s) + C O_{2} (g) \to N a_{2} C O_{3} (s) + H_{2} O (l)$ $2NaOH(s) + CO_2(g) -> Na_2CO_3(s) + H_2O(l)$