$B a C O_{3} + C (s) + H_{2} O (l) \to C O (g) + B a {(O H)}_{2}$ $BaCO_3 + C(s) + H_2O(l) -> CO(g) + Ba(OH)_2$ . If 394 g of $B a C O_{3}$ $BaCO_3$ are reacted with excess carbon and water, what mass of $B a {(O H)}_{2}$ $Ba(OH)_2$ should be produced?

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$B a C O_{3} + C (s) + H_{2} O (l) \to C O (g) + B a {(O H)}_{2}$ $BaCO_3 + C(s) + H_2O(l) -> CO(g) + Ba(OH)_2$ . If 394 g of $B a C O_{3}$ $BaCO_3$ are reacted with excess carbon and water, what mass of $B a {(O H)}_{2}$ $Ba(OH)_2$ should be produced?

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Answer 1 · 2017-03-05T17:45:21

You should expect 342.6 g of $B a {(O H)}_{2}$ $Ba(OH)_2$

Explanation:

$394 g o f B a C O_{3}$ $394 g of BaCO_3$ is 2.00 moles.

$\frac{394 g}{197.3 (\frac{g}{mol})} = 2.00$ $(394 g)/(197.3 (g/"mol")) = 2.00$ mol

Looking at the equation, you see that the coefficient of both $B a C O_{3}$ $BaCO_3$ and $B a {(O H)}_{2}$ $Ba(OH)_2$ is a one. So, the amount of $B a {(O H)}_{2}$ $Ba(OH)_2$ produced will be equal to the amount of $B a C O_{3}$ $BaCO_3$ reacted.

$2.00 mol B a {(O H)}_{2} \times 171.3 (\frac{g}{mol}) = 342.6 g$ $2.00 " mol " Ba(OH)_2 xx 171.3 (g/"mol") = 342.6 g$

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$B a C O_{3} + C (s) + H_{2} O (l) \to C O (g) + B a {(O H)}_{2}$ $BaCO_3 + C(s) + H_2O(l) -> CO(g) + Ba(OH)_2$ . If 394 g of $B a C O_{3}$ $BaCO_3$ are reacted with excess carbon and water, what mass of $B a {(O H)}_{2}$ $Ba(OH)_2$ should be produced?

1 Answer

Explanation:

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Math

Humanities

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BaCO3+C(s)+H2O(l)→CO(g)+Ba(OH)2BaCO_3 + C(s) + H_2O(l) -> CO(g) + Ba(OH)_2. If 394 g of BaCO3BaCO_3 are reacted with excess carbon and water, what mass of Ba(OH)2Ba(OH)_2 should be produced?

1 Answer

Explanation:

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$B a C O_{3} + C (s) + H_{2} O (l) \to C O (g) + B a {(O H)}_{2}$ $BaCO_3 + C(s) + H_2O(l) -> CO(g) + Ba(OH)_2$ . If 394 g of $B a C O_{3}$ $BaCO_3$ are reacted with excess carbon and water, what mass of $B a {(O H)}_{2}$ $Ba(OH)_2$ should be produced?